The bold sans-serif texts are the problems as given.

1 Stoichiometry

Problem 1

Consider the following set of reactions

4CH_{4} + O_{2} →^{1} 2C_{2}H_{6} + 2H_{2}O
2CH_{4} + O_{2} →^{2} 2CH_{3}OH
2CH_{3}OH + O_{2} →^{3} 2CH_{2}O + 2H_{2}O
2C_{2}H_{6} + 3O_{2} →^{4} 4CH_{2}O + 2H_{2}O

(a) Show, that there are 3 key components within the system! The number of key components is the number of species minus rank of the element species matrix.

N_{species} − Rg(\underline{B}) = N_{kc}

First build the element species matrix:

\underline{B} = ( \begin{matrix}& CH_{4} & O_{2} & C_{2}H_{6} & H_{2}O & CH_{3}OH & CH_{2}O \\ C & 1 & 0 & 2 & 0 & 1 & 1 \\ O & 0 & 2 & 0 & 1 & 1 & 1 \\ H & 4 & 0 & 6 & 2 & 4 & 2 \end{matrix} )

Calculate its rank using the Gauss method:

\underline{B} = ( \begin{matrix}1 & 0 & 2 & 0 & 1 & 1 \\ 0 & 2 & 0 & 1 & 1 & 1 \\ 4 & 0 & 6 & 2 & 4 & 2 \end{matrix} ) = ( \begin{matrix}1 & 0 & 2 & 0 & 1 & 1 \\ 0 & 2 & 0 & 1 & 1 & 1 \\ 0 & 0 & −2 & 2 & 0 & −2 \end{matrix} )

This shows we have 3 linearly independent equations, that is Rg(\underline{B}) = 3 = N_{nkc} (number of non-key compenents)

N_{species} − N_{nkc} = N_{kc} = 6 − 3 = 3

The number of key components is 3.

(b) For the chosen key components CH_{4}, O_{2} and C_{2}H_{6} the following changes in amount of substance have been measured:

Δn_{CH_{4}} = −8mol, Δn_{O_{2}} = −12mol, Δn_{C_{2}H_{6}} = 3mol

Calculate the changes in amount of substance of the non-key components using the element species matrix! This can be solved by using the relation:

\underline{B} · Δn = 0

= ( \begin{matrix}1 & 0 & 2 & 0 & 1 & 1 \\ 0 & 2 & 0 & 1 & 1 & 1 \\ 4 & 0 & 6 & 2 & 4 & 2 \end{matrix} ) · ( \begin{matrix}Δn_{CH_{4}} \\ Δn_{O_{2}} \\ Δn_{C_{2}H_{6}} \\ Δn_{H_{2}O} \\ Δn_{CH_{3}OH} \\ Δn_{CH_{2}O} \end{matrix} )

= ( \begin{matrix}1 & 0 & 2 & 0 & 1 & 1 \\ 0 & 2 & 0 & 1 & 1 & 1 \\ 4 & 0 & 6 & 2 & 4 & 2 \end{matrix} ) · ( \begin{matrix}−8mol \\ −12mol \\ 3mol \\ Δn_{H_{2}O} \\ Δn_{CH_{3}OH} \\ Δn_{CH_{2}O} \end{matrix} )

Now splitting this into equations of an equation system ...

−8mol + 2 · 3mol + Δn_{CH_{3}OH} + Δn_{CH_{2}O} = 0
2 · −12mol + Δn_{H_{2}O} + Δn_{CH_{3}OH} + Δn_{CH_{2}O} = 0
4 · −8mol + 6 · 3mol + 2Δn_{H_{2}O} + 4Δn_{CH_{3}OH} + 2Δn_{CH_{2}O} = 0

... one can solve the system. Rearranging gives:

Δn_{CH_{3}OH} + Δn_{CH_{2}O} = 2
Δn_{H_{2}O} + Δn_{CH_{3}OH} + Δn_{CH_{2}O} = 24
2Δn_{H_{2}O} + 4Δn_{CH_{3}OH} + 2Δn_{CH_{2}O} = 14

This can be written as a matrix again which makes it easier to solve it by using the Gauss method:

( \begin{matrix}0 & 1 & 1 & 2 \\ 1 & 1 & 1 & 24 \\ 2 & 4 & 2 & 14 \end{matrix} ) ⇝ ( \begin{matrix}1 & 1 & 1 & 24 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 0 & −34 \end{matrix} )

This gives:

2Δn_{CH_{3}OH} = −34mol ⇝ Δn_{CH_{3}OH} = −17mol
−17mol + Δn_{CH_{2}O} = 2 ⇝ Δn_{CH_{2}O} = 19mol
Δn_{H_{2}O} − 17mol + 19 = 24 ⇝ Δn_{H_{2}O} = 22mol

(c) Calculate the number of key reactions and give a set of them! To do this, we need to write the given reactions in matrix form. Note: The stoichiometric coefficients on the left are to be negated! The Gauss method is used again to calculate the rank of the matrix.

\underline{N}^{T} = ( \begin{matrix}& CH_{4} & O_{2} & C_{2}H_{6} & H_{2}O & CH_{3}OH & CH_{2}O \\ I & −4 & −1 & 2 & 2 & 0 & 0 \\ II & −2 & −1 & 0 & 0 & 2 & 0 \\ III & 0 & −1 & 0 & 2 & −2 & 2 \\ IV & 0 & −3 & −2 & 2 & 0 & 4 \end{matrix} )

= ( \begin{matrix}−4 & −1 & 2 & 2 & 0 & 0 \\ 0 & 1 & 2 & 2 & −4 & 0 \\ 0 & −1 & 0 & 2 & −2 & 2 \\ 0 & −3 & −2 & 2 & 0 & 4 \end{matrix} ) = ( \begin{matrix}−4 & −1 & 2 & 2 & 0 & 0 \\ 0 & 1 & 2 & 2 & −4 & 0 \\ 0 & 0 & 2 & 4 & −6 & 2 \\ 0 & 0 & 4 & 8 & −12 & 4 \end{matrix} )

= ( \begin{matrix}−4 & −1 & 2 & 2 & 0 & 0 \\ 0 & 1 & 2 & 2 & −4 & 0 \\ 0 & 0 & 2 & 4 & −6 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} )

This shows us, that Rg(\underline{N}^{T}) = 3 ⇝ hence we have 3 key reactions. We will build them from the last step of the Gauss elimination (negative ν on the left and positive on the right side).

4CH_{4} + O_{2} →^{1} 2C_{2}H_{6} + 2H_{2}O
4CH_{3}OH →^{2} O_{2} + 2C_{2}H_{6} + 2H_{2}O
6CH_{3}OH →^{3} 2C_{2}H_{6} + 4H_{2}O + 2CH_{2}O

(d) Calculate the extents of reaction with the changes in amount of substance given in (1b) and use them to determine again the changes in number of moles of the non-key components!

We know the relation \underline{N} · \underline{ξ_{j}} = \underline{Δn_{i}} and use the transposed matrix from (1c) skipping the 4th reaction which

had been found to be linearly dependent with reaction 3. Since \underline{N}^{T}^{T} = \underline{N} we can use it.

\underline{N} · \underline{ξ_{j}} = ( \begin{matrix}−4 & −2 & 0 \\ −1 & −1 & −1 \\ 2 & 0 & 0 \\ 2 & 0 & 2 \\ 0 & 2 & −2 \\ 0 & 0 & 2 \end{matrix} ) · ( \begin{matrix}ξ_{1} \\ ξ_{2} \\ ξ_{3} \end{matrix} ) = ( \begin{matrix}−8mol \\ −12mol \\ 3mol \\ Δn_{4} \\ Δn_{5} \\ Δn_{6} \end{matrix} )

Now writing this as an equation system in matrix form, we get:

( \begin{matrix}−4 & −2 & 0 & −8 \\ −1 & −1 & −1 & −12 \\ 2 & 0 & 0 & 3 \\ 2 & 0 & 2 & Δn_{4} \\ 0 & 2 & −2 & Δn_{5} \\ 0 & 0 & 2 & Δn_{6} \end{matrix} ) = ( \underline{N_{1}} ; \underline{N_{2}} )

Let’s now split the matrix into the upper three (N_{1} ⇝ key components) and the lower three (N_{2} ⇝ non-key components) matrices. Of which the latter one needs to be calculated. Since it is possibly easier to read, let’s write

\underline{N_{1}} = ( \begin{matrix}−4 & −2 & 0 & −8 \\ −1 & −1 & −1 & −12 \\ 2 & 0 & 0 & 3 \end{matrix} ) ⇝ \{ \begin{matrix}−4ξ_{1} − 2ξ_{2} & = & −8 \\ −1ξ_{1} − 1ξ_{2} − 1ξ_{3} & = & −12 \\ 2ξ_{1} & = & 3 \end{matrix}

One can easily see the first solution and the other solutions just depend on that first one:

2ξ_{1} = 3 ⇝ ξ_{1} = 1, 5
−4 · 1, 5 − 2ξ_{2} = −8 ⇝ ξ_{2} = 1
−1 · 1, 5 − 1 − 1ξ_{3} = −12 ⇝ ξ_{3} = 9, 5

Now it is easy to re-calculate the 3 non-key components again, using \underline{N_{2}}.

2 · 1, 5 + 2 · 9, 5 = Δn_{4} ⇝ Δn_{4} = 22mol
2 · 1 − 2 · 9, 5 = Δn_{5} ⇝ Δn_{5} = −17mol
2 · 9, 5 = Δn_{6} ⇝ Δn_{6} = 19mol

Since this is the same result as above, we are done ;-)

Problem 2

An unknown system with the species CH_{3}OH, O_{2}, CH_{2}O, H_{2}O and CO is investigated.

(a) For the so-called formula conversion Δn_{ij} = ν_{ij} the following formula is valid

\underline{B} · \underline{N} = 0

Assume the number of key reactions and key components being equal and give a set of key reactions! NOTE: Choose the stoichiometric coefficients of your chosen key components in such a way, that there is only one key component left in each equation! First let’s set up the element species matrix and determine its rank:

\underline{B} = ( \begin{matrix}& CH_{3}OH & O_{2} & CH_{2}O & H_{2}O & CO \\ C & 1 & 0 & 1 & 0 & 1 \\ O & 1 & 2 & 1 & 1 & 1 \\ H & 4 & 0 & 2 & 2 & 0 \end{matrix} )

= ( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 \\ 0 & 0 & −2 & 2 & −4 \end{matrix} )

Hence Rg(\underline{B}) = 3 and therefore we have 2 key components (N_{c} − Rg(\underline{B}) = N_{kc}). Now we build up \underline{N}, which yields:

\underline{B} · \underline{N} = 0

= \underline{B} · ( \begin{matrix}ν_{1,1} & ν_{1,2} \\ ν_{2,1} & ν_{2,2} \\ ν_{3,1} & ν_{3,2} \\ ν_{4,1} & ν_{4,2} \\ ν_{5,1} & ν_{5,2} \end{matrix} ) = ( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 4 & 0 & 2 & 2 & 0 \end{matrix} ) · ( \begin{matrix}ν_{1,1} & ν_{1,2} \\ 1 & 0 \\ ν_{3,1} & ν_{3,2} \\ ν_{4,1} & ν_{4,2} \\ 0 & 1 \end{matrix} )

The choice for ν_{2,1}, ν_{2,2}, ν_{5,1} and ν_{5,2} is done because of the condition that each key component shall only appear once per key reaction.

We now split the problem into two parts (one per reaction) - and we also use the matrix (\underline{B})we got by using the Gauss method (this allowed because the matrix is the same anyway, it just looks differently - sure, one could also take the original matrix, but this would require more work):

\underline{B} · \underline{N_{1}} = ( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 \\ 0 & 0 & −2 & 2 & −4 \end{matrix} ) · ( \begin{matrix}ν_{1,1} \\ 1 \\ ν_{3,1} \\ ν_{4,1} \\ 0 \end{matrix} ) = 0

\underline{B} · \underline{N_{2}} = ( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 \\ 0 & 0 & −2 & 2 & −4 \end{matrix} ) · ( \begin{matrix}ν_{1,2} \\ 0 \\ ν_{3,2} \\ ν_{4,2} \\ 1 \end{matrix} ) = 0

Now writing the first one as an equation system:

⇝ \{ \begin{matrix}ν_{1,1} + ν_{3,1} & = & 0 \\ ν_{4,1} & = & −2 \\ −2ν_{3,1} + 2ν_{4,1} & = & 0 ⇝ ν_{4,1} = ν_{3,1} \end{matrix} \} ⇝ ν_{1,1} = 2

This yields the first reaction equation now (check again if it’s balanced → it is!):

2CH_{2}O + 2H_{2}O → 2CH_{3}OH + O_{2}

Now do the same for the second reaction:

⇝ \{ \begin{matrix}ν_{1,2} + ν_{3,2} & = & −1 \\ ν_{4,2} & = & 0 \\ −2ν_{3,2} + 2ν_{4,2} & = & 4 ⇝ ν_{3,2} = −2 \end{matrix} \} ⇝ ν_{1,2} = 1

The resulting reaction equation looks like this:

2CH_{2}O → CH_{3}OH + CO

(b) Calculate the changes in amount of substance of CH_{3}OH, CH_{2}O and H_{2}O with given Δn_{O_{2}} = 10mol and Δn_{CO} = 2mol! Because the relation Δn_{ij} = ν_{ij} allows us to interchange the values for the stoichiometric coefficients ν and the change in amount of substance Δn, we can employ the already known relation:

\underline{B} · Δn = 0

( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 4 & 0 & 2 & 2 & 0 \end{matrix} ) · ( \begin{matrix}Δn_{1} \\ Δn_{2} \\ Δn_{3} \\ Δn_{4} \\ Δn_{5} \end{matrix} ) = ( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 4 & 0 & 2 & 2 & 0 \end{matrix} ) · ( \begin{matrix}Δn_{1} \\ 10mol \\ Δn_{3} \\ Δn_{4} \\ 2mol \end{matrix} ) = 0

Let’s now develop the equation system from the matrix:

( \begin{matrix}1 & 0 & 1 & 0 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 4 & 0 & 2 & 2 & 0 \end{matrix} ) ⇝ \{ \begin{matrix}Δn_{1} + Δn_{3} & = & −2mol \\ Δn_{1} + Δn_{3} + Δn_{4} & = & −22mol \\ 4Δn_{1} + 2Δn_{3} + 2Δn_{4} & = & 0mol \end{matrix}

Now writing this in matrix form while skipping column 2 and 5 (i.e. the given components Δn_{2} and Δn_{5}) again will make things easier:

\underline{M} = ( \begin{matrix}Column & Δn_{1} & Δn_{3} & Δn_{4} & Result \\ & 1 & 1 & 0 & −2 \\ & 1 & 1 & 1 & −22 \\ & 4 & 2 & 2 & 0 \end{matrix} ) = ( \begin{matrix}1 & 1 & 0 & −2 \\ 0 & 0 & 1 & −20 \\ 0 & −2 & 2 & 8 \end{matrix} )

⇝ \{ \begin{matrix}Δn_{4} & = & −20mol \\ Δn_{3} & = & −24mol \\ Δn_{1} & = & 22mol \end{matrix}

Problem 3

Consider the following set of reactions:

\begin{matrix}RK 1: & A_{1} + A_{2} → A_{3} + A_{4} \\ RK 2: & A_{2} + A_{3} → A_{4} + A_{5} \\ RK 3: & A_{2} + A_{5} → A_{4} + A_{6} \end{matrix}

There are three key reactions within the system. Calculate the changes in number of moles Δn_{2}, Δn_{4} and Δn_{5}, when the following changes have been measured:

Δn_{1} = −1mol
Δn_{3} = −2mol
Δn_{6} = 2mol

HINT: Set up the matrix of stoichiometric coefficients and make use of the extents of reaction. We already know the relation:

\underline{N} · \underline{ξ_{j}} = \underline{Δn_{i}}

\underline{N} = ( \begin{matrix}& I & II & III \\ A_{1} & −1 & 0 & 0 \\ A_{2} & −1 & −1 & −1 \\ A_{3} & 1 & −1 & 0 \\ A_{4} & 1 & 1 & 1 \\ A_{5} & 0 & 1 & −1 \\ A_{6} & 0 & 0 & 1 \end{matrix} ); \underline{ξ_{j}} = ( \begin{matrix}ξ_{1} \\ ξ_{2} \\ ξ_{3} \end{matrix} ); \underline{Δn_{i}} = ( \begin{matrix}Δn_{1} \\ Δn_{2} \\ Δn_{3} \\ Δn_{4} \\ Δn_{5} \\ Δn_{6} \end{matrix} )

( \begin{matrix}−1 & 0 & 0 \\ −1 & −1 & −1 \\ 1 & −1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & −1 \\ 0 & 0 & 1 \end{matrix} ) · ( \begin{matrix}ξ_{1} \\ ξ_{2} \\ ξ_{3} \end{matrix} ) = ( \begin{matrix}Δn_{1} \\ Δn_{2} \\ Δn_{3} \\ Δn_{4} \\ Δn_{5} \\ Δn_{6} \end{matrix} ) = ( \begin{matrix}−1mol \\ Δn_{2} \\ −2mol \\ Δn_{4} \\ Δn_{5} \\ 2mol \end{matrix} )

Writing this equation system as a matrix makes things easy again:

( \begin{matrix}A_{1} & −1 & 0 & 0 & −1 \\ A_{3} & 1 & −1 & 0 & −2 \\ A_{6} & 0 & 0 & 1 & 2 \end{matrix} ) ⇝ \{ \begin{matrix}ξ_{1} = 1 \\ ξ_{3} = 2 \\ ξ_{2} = 3 \end{matrix}

Now let’s calculate the missing (non-key) changes in amount of substance:

( \begin{matrix}A_{2} & −1 & −1 & −1 & Δn_{2} \\ A_{4} & 1 & 1 & 1 & Δn_{4} \\ A_{5} & 0 & 1 & −1 & Δn_{5} \end{matrix} ) ⇝ \{ \begin{matrix}−ξ_{1} − ξ_{2} − ξ_{3} & = Δn_{2} = & −6mol \\ ξ_{1} + ξ_{2} + ξ_{3} & = Δn_{4} = & 6mol \\ ξ_{2} − ξ_{3} & = Δn_{5} = & 1mol \end{matrix}

Problem 4

The following species are involved in a set of reactions:

\begin{matrix}A_{1} = C_{6}H_{6}; & A_{2} = Cl_{2} \\ A_{3} = C_{6}H_{5}Cl; & A_{4} = HCl \\ A_{5} = C_{6}H_{4}Cl_{2}; & A_{6} = C_{6}H_{3}Cl_{3} \end{matrix}

Calculate the changes in number of moles Δn_{1}, Δn_{2} and Δn_{4}, when the following changes have been measured:

\begin{matrix}Δn_{3} = & −2mol \\ Δn_{5} = & 1mol \\ Δn_{6} = & 2mol \end{matrix}

HINT: Use the element-species matrix. The number of key components has \underline{not} to be calculated. Just set up the element-species matrix (\underline{B}):

\underline{B} = ( \begin{matrix}& A_{1} & A_{2} & A_{3} & A_{4} & A_{5} & A_{6} \\ C & 6 & 0 & 6 & 0 & 6 & 6 \\ H & 6 & 0 & 5 & 1 & 4 & 3 \\ Cl & 0 & 2 & 1 & 1 & 2 & 3 \end{matrix} ); \underline{Δn_{i}} = ( \begin{matrix}Δn_{1} \\ Δn_{2} \\ Δn_{3} \\ Δn_{4} \\ Δn_{5} \\ Δn_{6} \end{matrix} ) = ( \begin{matrix}Δn_{1} \\ Δn_{2} \\ −2mol \\ Δn_{4} \\ 1mol \\ 2mol \end{matrix} )

The following formula is to be used:

\underline{B} · Δn = 0

Writing this as an equation system an rearranging gives:

⇝ \{ \begin{matrix}6Δn_{1} & = & −6 & ⇝ & Δn_{1} = & −1mol \\ 6 · −1 + Δn_{4} & = & 0 & ⇝ & Δn_{4} = & 6mol \\ 2Δn_{2} & = & −12 & ⇝ & Δn_{2} = & −6mol \end{matrix}

2 Thermodynamics

Problem 1

During the production of synthesis gas, the conversion to water gas is a side reaction:

CO(g) + H_{2}O(g) ⇀ ↽ CO_{2}(g) + H_{2}(g);

the equilibrium is adjustable by the reaction conditions. The reaction is carried out at constant temperature in a closed vessel until equilibrium is reached. Following thermodynamic data is given:

1. Standard molar heats of formation:

\begin{aligned}CO(g): & ΔH_{f}^{0} & = & −111\frac{kJ}{mol} \\ CO_{2}(g): & ΔH_{f}^{0} & = & −393\frac{kJ}{mol} \\ H_{2}O(fl): & ΔH_{f}^{0} & = & −283\frac{kJ}{mol} \end{aligned}

2. Vaporization of water:

H_{2}O(fl) → H_{2}O(g): ΔH_{vap}^{0} = − + 41\frac{kJ}{mol}

3. Constant pressure heat capacities:

Δc_{p} = A + BT + CT^{2} + DT^{3} with [T] = K and [Δc_{p}] = \frac{cal}{mol·K}
1cal = 4, 184J
Species  Coefficients
A  B  C  D
CO_{2}  4, 728 1, 754 · 10^{−2} −1, 338 · 10^{−5}  4, 097 · 10^{−9}
CO  7, 373 −0, 307 · 10^{−2}  6, 662 · 10^{−6} −3, 037 · 10^{−9}
H_{2}O  7, 701 4, 595 · 10^{−4}  2, 521 · 10^{−6} −0, 859 · 10^{−9}
H_{2}  6, 483 2, 215 · 10^{−3} −3, 298 · 10^{−6}  1, 826 · 10^{−9}

(a) Calculate the standard enthalpy change on reaction ΔH_{R}^{0} using the given thermodynamical data! (Consider the molecules’ state of aggregation!) Using the following relation, we can easily calculate the reaction enthalpy:

ΔH_{R}^{0} = ∑_{i=1}^{N} ν_{i}ΔH_{f,i}
= \underbrace{−(−111)}_{CO(g)} \underbrace{−(−283 + 41)}_{H_{2}O(g)} − \underbrace{393}_{CO_{2}(g)} + \underbrace{0}_{H_{2}(g)} [\frac{kJ}{mol} ]
= −40\frac{kJ}{mol}

Hence the reaction is exothermic (i.e. it produces heat).

(b) In which direction is the equilibrium shifted in response to the following changes within the system?

1. Feed of further CO

2. Removal of a fixed amount of CO_{2} out of the vessel

3. Feed of further H_{2}

4. Addition of a catalyst

5. Increase in temperature

6. Increase in pressure The solution can be easily determined by using the different K_{x} and some basic chemical knowledge:

1. → In favor for products, use K_{c}

2. → In favor for products, use K_{c}

3. ← In favor for educts, use K_{c}

4. ⇀ ↽ No change. Catalysts only speed up or slow down reactions. The equilibrium is never affected!

5. ← In favor for educts, use K_{a}. The reaction is exothermic and therefore produces heat itself. Adding heat shifts the equilibrium!

6. ⇀ ↽ No change, use K_{p}. There is no change in volume during the reaction. Hence the partial pressures of all species should be the same and the quotient remains constant.

(c) Calculate the enthalpy change on reaction ΔH_{R,1000} for a temperature of 1000 K using the given thermodynamical data! We will have to calculate the reaction enthalpy at a given temperature upon the basis of the already calculated standard reaction enthalpy:

ΔH_{R}(T) = ΔH_{R}^{0} + ∫_{T_{0}}^{T} Δc_{p}(T)dT

Given the function for Δc_{p} we have:

ΔH_{R,1000} = −40\frac{kJ}{mol} + ∫_{298}^{1000} (A_{R} + B_{R} · T + C_{R} · T^{2} + D_{R} · T^{3}) dT
= −40\frac{kJ}{mol} + (A_{R} + B_{R} · T |_{298}^{1000} + C_{R} · T^{2}|_{298}^{1000} + D_{R} · T^{3}|_{298}^{1000})

For the constants A, B, C and D, given for each species, we apply the following formula to get the overall constants A_{R}, B_{R}, C_{R} and D_{R}:

Z_{R} = ∑_{i=1}^{N} ν_{i}Z_{i} where Z = [A, B, C, D]

Let’s calculate:

A_{R} = −7, 373 − 7, 701 + 4, 728 + 6, 483
= −3, 863
B_{R} = −(−0, 307 · 10^{−2}) − 4, 595 · 10^{−4} + 1, 754 · 10^{−2} + 2, 215 · 10^{−3}
= 0, 0223655
C_{R} = −6, 662 · 10^{−6} − 2, 521 · 10^{−6} − 3, 298 · 10^{−6} + 1, 338 · 10^{−5}
= −2, 586110^{−5}
D_{R} = 3, 037 · 10^{−9} + 0, 859 · 10^{−9} + 1, 826 · 10^{−9} + 4, 097 · 10^{−9}
= 9, 819 · 10^{−9}

Now it is possible to calculate the final result (partially without units):

∫_{298}^{1000} (A_{R} + B_{R} · T + C_{R} · T^{2} + D_{R} · T^{3}) dT
= −3, 863 · T |_{298}^{1000} + 0, 0223655 · \frac{1}{2} T^{2}|_{298}^{1000} + −2, 5861 · 10^{−5} · \frac{1}{3} T^{3}|_{298}^{1000}

+9, 819 · 10^{−9} · \frac{1}{4} T^{4}|_{298}^{1000}

= −3, 863 · (1000 − 298) + \frac{1}{2} (0, 0223655 · (1000^{2} − 298^{2})) + \frac{1}{3} (−2, 5861 · 10^{−5} · (1000^{3} − 298^{3}))

+\frac{1}{4} (9, 819 · 10^{−9} · (1000^{4} − 298^{4}))

= −2711, 826 + 10189, 67707 − 8392, 208349 + 2435, 391472 = 1521, 034193\frac{cal}{mol}
= 1, 521034193\frac{kcal}{mol} = 6, 364007065\frac{kJ}{mol}

Now plugging this into the above equation:

ΔH_{R,1000} = ΔH_{R}^{0} + ∫_{298}^{1000} Δc_{p}(T)dT
= −40\frac{kJ}{mol} + 6, 364007065\frac{kJ}{mol}
= −33, 635992935\frac{kJ}{mol}

Problem 2

The heat tonality of the experimentally not realizable formation of methanol from the elements (see the equation) cannot be measured directly:

C(s) + 2H_{2}(g) + \frac{1}{2}O_{2}(g) →^{ΔH_{R}^{0}} CH_{3}OH(fl),

but the heats of combustion of carbon, hydrogen and methanol can be measured calorimetrically:

C(s) + O_{2}(g) →^{ΔH_{R,1}^{0}} CO_{2}(g)
H_{2}(g) + \frac{1}{2}O_{2}(g) →^{ΔH_{R,2}^{0}} H_{2}O(fl)
CH_{3}OH(fl) + \frac{3}{2}O_{2}(g) →^{ΔH_{R,3}^{0}} 2H_{2}O(fl) + CO_{2}(g)

(a) Calculate the enthalpy change on reaction ΔH_{R}^{0} of the formation of methanol from the elements! NOTE: Represent the methanol’s formation reaction by a linear combination of the three combustion processes. We have know formula to calculate the enthalpy change on reaction from the formation enthalpies:

ΔH_{R}^{0} = ∑_{i=1}^{N} ν_{i}ΔH_{f,i}^{0}

Furthermore we know, that ΔH_{f,i}^{0} for any element is zero. Let’s use this, since it will cancel out several unknowns (following the sums in form ν_{i}ΔH_{f,i}^{0}):

Form. from elem.: ΔH_{R}^{0} = −1 · 0 − 2 · 0 − \frac{1}{2} · 0 + 1 · ΔH_{f,CH_{3}OH}^{0}
RK1: ΔH_{R,1}^{0} = −1 · 0 − 1 · 0 + 1 · ΔH_{f,CO_{2}}^{0} ⇝ ΔH_{f,CO_{2}}^{0} = −393\frac{kJ}{mol}
RK2: ΔH_{R,2}^{0} = −1 · 0 − \frac{1}{2} · 0 + 1 · ΔH_{f,H_{2}O}^{0} ⇝ ΔH_{f,H_{2}O}^{0} = −285\frac{kJ}{mol}
RK3: ΔH_{R,3}^{0} = −1 · ΔH_{f,CH_{3}OH}^{0} − \frac{3}{2} · 0 + 2 · ΔH_{f,H_{2}O}^{0} + 1 · ΔH_{f,CO_{2}}^{0}

For the third equation we now know the two remaining unknowns ΔH_{f,H_{2}O}^{0} and ΔH_{f,CO_{2}}^{0}, hence the equation look now like this:

RK3:  ΔH_{R,3}^{0}  = −1 · ΔH_{f,CH_{3}OH}^{0} + 2 · −285\frac{kJ}{mol} + 1 · −393\frac{kJ}{mol}

= −ΔH_{f,CH_{3}OH}^{0} − 570\frac{kJ}{mol} − 393\frac{kJ}{mol} = −ΔH_{f,CH_{3}OH}^{0} − 963\frac{kJ}{mol}

⇝ ΔH_{R,3}^{0} + 963\frac{kJ}{mol} = −ΔH_{f,CH_{3}OH}^{0} = −724\frac{kJ}{mol} + 963\frac{kJ}{mol}
⇝ ΔH_{f,CH_{3}OH}^{0}  = −239\frac{kJ}{mol}

Now we’ll have to use the equation (”Form. from elem.”) from above to get the result for ΔH_{R}^{0} of the formation of methanol from the elements:

ΔH_{R}^{0} = −1 · 0 − 2 · 0 − \frac{1}{2} · 0 + 1 · ΔH_{f,CH_{3}OH}^{0}
ΔH_{R}^{0} = ΔH_{f,CH_{3}OH}^{0} = −239\frac{kJ}{mol}

(b) Is the reaction in problem 2a exothermic or endothermic? methanol from the elements! NOTE: Represent the methanol’s formation reaction by a linear combination of the three combustion processes. The result (ΔH_{R}^{0} = −239\frac{kJ}{mol}) clearly shows, that this reaction is exothermic.

Problem 3

The temperature dependence of the balanced reaction:

SO_{2}(g) + \frac{1}{2}O_{2}(g) ⇀ ↽^{ΔH_{R}^{0}} SO_{3}(g), ΔH_{R}^{0} = −99\frac{kJ}{mol}

is of decisive importance for the production of sulphuric acid. The standard molar Gibbs free energies of formation are known:

SO_{2} : ΔG_{f}^{0} = −300\frac{kJ}{mol}
SO_{3} : ΔG_{f}^{0} = −371\frac{kJ}{mol}

(a) Calculate the standard Gibbs free energy change on reaction ΔG_{R}^{0}! We use the following relation:

ΔG_{R}^{0} = ∑_{i=1}^{N} ν_{i}ΔG_{f,i}

= −1 · ΔG_{f,SO_{2}} + (−\frac{1}{2}) · ΔG_{f,O_{2}} + 1 · ΔG_{f,SO_{3}}
= −(−300\frac{kJ}{mol}) − (\frac{1}{2} · \underbrace{0\frac{kJ}{mol}}_{zero for elements!}) + (−371\frac{kJ}{mol})

= −71\frac{kJ}{mol}

(b) Calculate the standard activity equilibrium ratio K_{a}^{0}!

G_{R}^{0} = −RT · ln K_{a}
⇝ −\frac{G_{R}^{0}}{RT} = ln K_{a}
⇝ K_{a} = exp (−\frac{G_{R}^{0}}{RT} )

⇝ K_{a}^{0} = exp (−\frac{G_{R}^{0}}{R · 298K} ) = exp (−\frac{−71\frac{kJ}{mol}}{8, 314\frac{J}{mol·K} · 298K} ) = e^{28,6570884721009} = 2790071872271, 74

This means, that the standard activity equlibrium shifts the balance to the product’s side:

K_{a}^{0} = \frac{[SO_{3}]}{[SO_{2}] [O_{2}]} = 2, 79007187227174 · 10^{12}

(c) Calculate the activity equilibrium ratio K_{a} for the following temperatures:

T = 713K; 733K; 753K; 773K; 873K; 973K; 1073K; 1173K

Using the relation from (b), we could easily determine K_{a} for changing T if we knew G_{R}(T):

K_{a} = exp (−\frac{G_{R}^{0}}{RT} ) = exp (−\frac{G_{R}(T)}{RT})

This is not the case, hence we have to use another relation:

\frac{d ln K_{a}}{dT} = \frac{ΔH_{R}^{0}}{RT^{2}} = ⇝ ∫_{K_{a}^{0}}^{K_{a}} d ln K_{a} = ∫_{T_{0}}^{T} \frac{ΔH_{R}^{0}}{RT^{2}} dT = ln K_{a} − ln K_{a}^{0}

Rearranging it gives:

ln K_{a} = ln K_{a}^{0} + ∫_{T_{0}}^{T} \frac{ΔH_{R}^{0}}{RT^{2}} dT

Moving the constants out of the integral:

ln K_{a}(T) = ln K_{a}^{0} + \frac{ΔH_{R}^{0}}{R} · ∫_{T_{0}}^{T} \frac{dT}{T^{2}} = ln K_{a}^{0} + \frac{ΔH_{R}^{0}}{R} · − \frac{1}{T} |_{T_{0}}^{T}

Now plugging all known variables in and calculating for each given temperature yields:

ln K_{a}(T) = 28, 6570884721009 + \frac{−99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · − \frac{1}{T} |_{298}^{T}
ln K_{a}(713K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{713K} − \frac{1}{298K} = 5, 39935029229595
ln K_{a}(733K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{733K} − \frac{1}{298K} = 4, 9436685107031
ln K_{a}(753K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{753K} − \frac{1}{298K} = 4, 51219293264772
ln K_{a}(773K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{773K} − \frac{1}{298K} = 4, 10304468075303
ln K_{a}(873K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{873K} − \frac{1}{298K} = 2, 33850496897352
ln K_{a}(973K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{973K} − \frac{1}{298K} = 0, 936666122924645
ln K_{a}(1073K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{1073K} − \frac{1}{298K} = −0, 203879368781479
ln K_{a}(1173K) = 28, 6570884721009 + \frac{99000\frac{J}{mol}}{8, 314\frac{J}{mol·K}} · \frac{1}{1173K} − \frac{1}{298K} = −1, 14995845098955
K_{a}(713K) = exp (5, 39935029229595) = 221, 262613469784
K_{a}(733K) = exp (4, 9436685107031) = 140, 283939879604
K_{a}(753K) = exp (4, 51219293264772) = 91, 1214227140278
K_{a}(773K) = exp (4, 10304468075303) = 60, 5242844733766
K_{a}(873K) = exp (2, 33850496897352) = 10, 365727887841
K_{a}(973K) = exp (0, 936666122924645) = 2, 55146096592772
K_{a}(1073K) = exp (−0, 203879368781479) = 0, 815560747340286
K_{a}(1173K) = exp (−1, 14995845098955) = 0, 316649925596806

This shows us, that the equilibrium shifts to the left (educts) when adding heat to the exothermic reaction.

Problem 4

Consider the following gas-phase reaction in equilibrium

RK1 : A_{1} + 2A_{2} ⇀ ↽ 3A_{3}

with standard heat of reaction ΔH_{R,1}^{0}, A_{1} being H_{2}, and all species behaving like ideal gases.

(a) Calculate the standard heat of reaction ΔH_{R,1}^{0} when there is given a second reaction

RK2 : A_{1} + A_{4} ⇀ ↽ A_{2}

with the standard heat of reaction ΔH_{R,2}^{0} = 100\frac{kJ}{mol} and the following standard molar heats of formation for A_{3} and A_{4}:

ΔH_{f,3}^{0} = 20\frac{kJ}{mol}; ΔH_{f,4}^{0} = 50\frac{kJ}{mol}

We need to employ the already known relation:

ΔH_{R}^{0} = ∑_{i=1}^{N} ν_{i} · ΔH_{f,i}^{0}

This is a simple linear equation system. We know several of the variables, so let’s start:

(RK2) : ΔH_{R,2}^{0} = 100\frac{kJ}{mol} = \underbrace{−1 · 0}_{A_{1}} + \underbrace{(−1) · 50\frac{kJ}{mol}}_{A_{4}} +1 · ΔH_{f,2}^{0} ⇝ ΔH_{f,2}^{0} = 150\frac{kJ}{mol}

So, why is ΔH_{f,1}^{0} = 0? Because H_{2} is an element and this is one condition for elements! Now we can calculate ΔH_{R,1}^{0}:

ΔH_{R,1}^{0} = \underbrace{−1 · 0}_{A_{1}} + \underbrace{(−2) · 150\frac{kJ}{mol}}_{A_{2}} + \underbrace{3 · 20\frac{kJ}{mol}}_{A_{3}} = (−300 + 60)\frac{kJ}{mol} = −240\frac{kJ}{mol}

(b) Does it favor the formation of educts or products in the reaction (RK1), when the partial pressures p_{1} and p_{2} of the species A_{1} and A_{2} are both 25% of the total pressure p? For the following relation

K_{p} = ∏_{i=1}^{N} p_{i}^{ν_{i}},

and under the condition, that the overall pressure must be 100% (p = p_{1} + p_{2} + p_{3} = 1 = 0, 25 + 0, 25 + p_{3} ⇝ p_{3} = 0, 5 = 50%) we get

K_{p} = p_{3}^{ν_{3}} · p_{1}^{ν_{1}} · p_{2}^{ν_{2}} = p_{3}^{3} · p_{1}^{−1} · p_{2}^{−2} = \frac{p_{3}^{3}}{p_{1}^{1} · p_{2}^{2}} = \frac{0, 5^{3}}{0, 25^{1} · 0, 25^{2}} = \frac{0, 5^{3}}{0, 25^{3}} = 8.

Hence the equilibrium shifts in favor for the products (i.e. →).

(c) In which direction is the equilibrium shifted in the case the reaction is endothermic and a catalyst is added? First of all, the reaction is exothermic. So this question needs not to be answered. And anyway, the addition of a catalyst does not influence the equilibrium. It just speeds up or slows down the reaction.

Problem 5

The isomerisation reaction

A_{1} ⇀ ↽ A_{2} ΔH_{f,1}^{0} = 18, 98\frac{kJ}{mol} ΔH_{f,2}^{0} = 17, 93\frac{kJ}{mol}

is in equilibrium at standard conditions, the equilibrium conversion X_{1}^{∗} is 0,45.

(a) Calculate K_{c}^{0} and K_{a}^{0} using the stoichiometric table. The stoichiometric table can be found in the Summary or can be built. First of all we assume a batch system.

ν_{1}A_{1} + ν_{2}A_{2} = 0
Species n_{i0} Δn_{i} = −\frac{ν_{i}}{ν_{k}} n_{k0}X_{k}  n_{i} = n_{i0} + Δn_{i}
A_{1}  n_{10}  −n_{10}X_{1}  n_{1} = n_{10} − n_{10}X_{1}
A_{2}  n_{20}  −\frac{ν_{2}}{ν_{1}} n_{10}X_{1}  n_{2} = n_{20} − \frac{ν_{2}}{ν_{1}} n_{10}X_{1}
Sum n_{sum0}  −  nsum = n_{sum0} − (1 + \frac{ν_{2}}{ν_{1}} )n_{10}X_{1}

We know, that n_{20} = 0 before the reaction starts. From the power law we know K_{a}^{0} = ∏_{i=1}^{N} c_{i}^{ν_{i}} and c_{i} = \frac{n_{i}}{V}, hence:

K_{a}^{0} = c_{1}^{−1} · c_{2}^{1} = \frac{c_{2}}{c_{1}} = \frac{n_{2}}{V} · \frac{V}{n_{1}} = \frac{n_{2}}{n_{1}}

Now from the stoichiometric table and the condition c_{20} = n_{20} = 0 we know:

K_{a}^{0} = \frac{n_{2}}{n_{1}} = \frac{n_{20} − \frac{ν_{2}}{ν_{1}} · n_{10} · X_{1}}{n_{10} − n_{10} · X_{1}} = \frac{−\frac{ν_{2}}{ν_{1}} · n_{10} · X_{1}}{(1 − X_{1}) · n_{10}} = \frac{−\frac{ν_{2}}{ν_{1}} · X_{1}}{1 − X_{1}} = \frac{1 · X_{1}}{1 − X_{1}} = \frac{0, 45}{1 − 0, 45} = \frac{0, 45}{0, 55} = 0, 81\overline{81}

(b) Calculate the standard heat of reaction. We already know how to do that:

ΔH_{R}^{0} = ∑_{i=1}^{N} ν_{i}ΔH_{f,i} = −1 · 18, 98\frac{kJ}{mol} + 1 · 17, 93\frac{kJ}{mol} = −1, 05\frac{kJ}{mol}

(c) Calculate K_{a} at a temperature of 400 K.

ln K_{a}(T) = ln K_{a}^{0} + ∫_{T_{0}}^{T} \frac{ΔH_{R}^{0}}{RT^{2}} dT

ln K_{a}(400K) = ln 0, 81\overline{81} + ∫_{298K}^{400K} \frac{−1, 05\frac{kJ}{mol}}{8, 3145\frac{J}{mol·K} · T^{2}} dT

Moving the constants out of the integral we get:

ln K_{a}(400K) = ln 0, 81\overline{81} + \frac{−1050\frac{J}{mol}}{8, 3145\frac{J}{mol·K}} ∫_{298K}^{400K} \frac{1}{T^{2}} dT = ln 0, 81\overline{81} + \frac{−1050\frac{J}{mol}}{8, 3145\frac{J}{mol·K}} [−\frac{1}{T} ]_{298K}^{400K}

= ln 0, 81\overline{81} + \frac{−1050\frac{J}{mol}}{8, 3145\frac{J}{mol·K}} (−\frac{1}{400K} − (−\frac{1}{298K} ))
= ln 0, 81\overline{81} − 126, 285405\frac{1}{K} (\frac{1}{298K} − \frac{1}{400K} ) = −0, 3087337099
K_{a}(400K) = 0, 7343763011

3 Microkinetics

Problem 1

In a laboratory experiment, 500ml of a solution of 2kmol/m^{3} ethylene oxide (A_{1}) in water was mixed with 500ml of water (A_{2}) containing 0,9% (w/w) sulfuric acid:

A_{1} + A_{2} → A_{3}

The concentration of ethylene glycol (A_{3}) was recorded as a function of time. From this data determine the reaction rate constant.

Time Concentration of A_{3}
0,0  0,000
0,5  0,145
1,0  0,270
1,5  0,376
2,0  0,467
3,0  0,610
4,0  0,715
6,0  0,848
10,0  0,957

From the definition of the rate of reaction we get:

r = \frac{1}{ν_{i}} · \frac{dc_{i}}{dt} = k · c_{1}c_{2} = \frac{dc_{3}}{dt} = \frac{c_{3}(t_{0} + Δt) − c_{3}(t_{0})}{Δt}

We can neglect c_{2} because water is in excess.

r = k · c_{1}

From the stoichiometric table we know:

c_{i} = c_{i0} − \frac{ν_{i}}{ν_{k}} c_{k0}X_{k} ⇝ c_{1} = c_{10} − c_{10}X_{1}

From c_{30} = 0 we can derive:

c_{3} = c_{30} − \frac{ν_{3}}{ν_{1}} c_{10}X_{1} = c_{10}X_{1} ⇝ c_{1} = c_{10} − c_{3}

r = k · (c_{10} − c_{3})^{κ} ⇝ \underbrace{log r}_{y} = \underbrace{log k}_{a} + \underbrace{κ}_{b} \underbrace{log(c_{10} − c_{3})}_{x}

The value for r can be calculated at different points in time, since we have the data for c_{3} and c_{10} from the given table. Plotting log r against log(c_{10} − c_{3}) in an X/Y-diagram allows us to estimate log k (i.e. a) and κ (i.e. b). Where a is the distance from the X-axis to the curve and b is the (mean-)inclination of the curve. Note: There is also the integral method available, but I believe that in this case it is better to use the differential method.

Problem 2

Consider the following reaction scheme, where A^{∗} is an activated molecule of the species A.

2A ⇀ ↽_{2}^{1} A^{∗} + A

A^{∗}  →^{3} product

Derive the formula for the rate of the third reaction as a function of the concentration of A and the rate constants by

(a) formulating equations for the reaction rate of each partial reaction assuming elementary reaction, Reaction rates:

r_{1} = k_{1}c_{A}^{2}
r_{2} = k_{2}c_{A}c_{A^{∗}}
r_{3} = k_{3}c_{A^{∗}}

(b) formulating an equation for the production rate R_{A^{∗}}, Production rate for every species:

R_{A^{∗}} = r_{1} − r_{2} − r_{3} = k_{1}c_{A}^{2} − k_{2}c_{A}c_{A^{∗}} − k_{3}c_{A^{∗}}

(c) formulating an expression for the concentration of A^{∗}, assuming R_{A^{∗}} = 0 and Concentration of A^{∗}:

0 = k_{1}c_{A}^{2} − k_{2}c_{A}c_{A^{∗}} − k_{3}c_{A^{∗}}
k_{1}c_{A}^{2} = k_{2}c_{A}c_{A^{∗}} + k_{3}c_{A^{∗}}
k_{1}c_{A}^{2} = c_{A^{∗}} · (k_{2}c_{A} + k_{3})
c_{A^{∗}} = \frac{k_{1}c_{A}^{2}}{k_{2}c_{A} + k_{3}}

(d) formulating the formula for r_{3}. Reaction rate for reaction 3:

r_{3} = k_{3} · \frac{k_{1}c_{A}^{2}}{k_{2}c_{A} + k_{3}}

Problem 3

Consider the following set of homogeneous reactions

A_{1} + 2A_{2} →^{1} A_{3} A_{4} ; ↗^{2} ; ↘_{3} ; A_{5}

with the reaction rates

r_{1} = 120\frac{kmol}{m^{3} · s}; r_{2} = 5r_{3}

(a) Formulate the equations for the reaction rate of each partial reaction assuming elementary reaction! Use the power law:

r_{1} = k_{1}c_{1}c_{2}^{2} = 120\frac{kmol}{m^{3} · s}
r_{2} = k_{2}c_{3} = 5r_{3}
r_{3} = k_{3}c_{3} = \frac{1}{5}r_{2}

(b) Calculate the net rates R_{i} for all species assuming the concentration c_{3} of species A_{3} being constant! Calculate the net rate using R_{i} = ∑_{j=1}^{M} ν_{ij} · r_{j}

R_{1} = −r_{1}
R_{2} = −2r_{1}
R_{3} = r_{1} − r_{2} − r_{3}
R_{4} = r_{2}
R_{5} = r_{3}

Problem 4

The description of a homogeneous catalyzed reaction

A_{1} + A_{2} ⇀ ↽_{2}^{1} A_{12} →^{3} A_{3} + A_{2}

follows enzyme kinetics, in which the homogeneous catalyst is represented by A_{12}.

(a) Form the equations for the reaction rate of every partial reaction assuming elementary reaction!

r_{1} = k_{1}c_{1}c_{2}
r_{2} = k_{2}c_{12}
r_{3} = k_{3}c_{12}

(b) Give the net rates of the changes in amount of substance of A_{1}, A_{2}, A_{3} and A_{12}!

R_{1} = −r_{1} = −k_{1}c_{1}c_{2}
R_{2} = −r_{1} + r_{2} + r_{3} = k_{2}c_{12} − k_{1}c_{1}c_{2} + k_{3}c_{12}
R_{3} = k_{3}c_{12}
R_{12} = r_{1} − r_{2} − r_{3} = k_{1}c_{1}c_{2} − k_{2}c_{12} − k_{3}c_{12}

(c) Determine the reaction rate of the formation of product assuming a (not?^{1}) neglectable amount of A_{2} being bound in the complex A_{12} during the reaction and

1. step 3 to be rate determining!

2. the steady state hypothesis to be valid for A_{12}!

1”not” was in the original script, though it makes no sense. The text without ”not” makes sense.