Chapter 17, Problem 17.85QA

Interpretation Introduction

To find:

a)  Assign oxidation number to carbon and hydrogen in the reactants and products.

b)  Find the standard free energy changes in the two reactions and the overall $∆G°$ for the formation of $H_2+ C{O}_2$ from methane and steam.

Answer to Problem 17.85QA

Solution:

a)  The oxidation numbers assigned to carbon and hydrogen in the reactants and products are as follows:

$-4+1 +1-2 +2-2 0$

$C{H}_4 \left(g\right)+H_{2}O \left(g\right)\to CO \left(g\right)+3H_2 \left(g\right)$

$+2-2 +1-2 0 +4-2$

$CO \left(g\right)+H_{2}O \left(g\right)\to H_2 \left(g\right)+ C{O}_2 \left(g\right)$

b)  The standard free energy changes in the two reactions:

$C{H}_4 \left(g\right)+H_{2}O \left(g\right)\to CO \left(g\right)+3H_2 \left(g\right) ∆G_{rxn}°=142.2 kJ$

$CO \left(g\right)+H_{2}O \left(g\right)\to H_2 \left(g\right)+ C{O}_2 \left(g\right) ∆G_{rxn}°=-28.6 kJ$

Overall $∆G°$ for the formation of $H_2+ C{O}_2$ from methane and steam:

$C{H}_4 \left(g\right)+2H_{2}O \left(g\right)\to {CO}_2 \left(g\right)+4H_2 \left(g\right) ∆G_{overall}°=113.6 kJ$

Explanation of Solution

1) Concept:

To assign the oxidation number of carbon and hydrogen in the reactant and product sides, we will use the guidelines of oxidation number in section $8.6$ in the text. We know that the oxidation numbers of oxygen and hydrogen in compounds usually are $-2$ and $+1$ respectively.

We will calculate the oxidation number of $C$ from the oxidation number of $O$ and $H$.

To calculate the standard free energy of reaction, we can use the standard heat of formation of products and reactants. From this, we can calculate the overall $∆G°$ of the reaction using Hess’s law.

2) Formula:

i) ${∆G_{rxn}}^0=\sum n_{product}∆{G_{f, product}}^0+ \sum n_{reactant}∆{G_{f, reactant}}^0$

ii) ${∆G_{overall}}^0={∆G_{{rxn}_1}}^0+ {∆G_{{rxn}_2}}^0$

3) Given:

i) ${∆G_f}^0 C{H}_4\left(g\right)= -50.8 kJ/mol$      

ii) ${∆G_f}^0 CO \left(g\right)= -137.2 kJ/mol$

iii) ${∆G_f}^0 C{O}_2\left(g\right)= -394.4 kJ/mol$

iv) ${∆G_f}^0 H_2\left(g\right)= 0.0 kJ/mol$

v) ${∆G_f}^0 H_{2}O \left(g\right)= -228.6 kJ/mol$

The standard free energy of formation values are taken from Appendix 4 of the text.

4) Calculations:

a) We assign the oxidation number to carbon and hydrogen in reaction by using rules for assigning the oxidation numbers to atoms.

The oxidation number of $C$ is unknown, so we call it x. The oxidation numbers of $C$ in $C{H}_4$, $C{O}_2$ and $CO$ are

$C{H}_4$ $x+4\left(+1\right)=0 x= -4$

$CO$ $x+1\left(-2\right)=0 x= +2$

$C{O}_2$ $x+2\left(-2\right)=0 x= +4$

The oxidation number of $H_2$ is $0$ because it is in elemental state.

The oxidation numbers to carbon and hydrogen in the reactants and products:

$-4+1 +1-2 +2-2 0$

$C{H}_4 \left(g\right)+H_{2}O \left(g\right)\to CO \left(g\right)+3H_2 \left(g\right)$

$+2-2 +1-2 0 +4-2$

$CO \left(g\right)+H_{2}O \left(g\right)\to H_2 \left(g\right)+ C{O}_2 \left(g\right)$

b) We can calculate the standard free energy of the first reaction by using the standard free energy of formation values taken from Appendix 4 of the text and the formula.

$C{H}_4 \left(g\right)+H_{2}O \left(g\right)\to CO \left(g\right)+3H_2 \left(g\right)$

${∆G_{rxn}}^0=\sum n_{product}∆{G_{f, product}}^0+ \sum n_{reactant}∆{G_{f, reactant}}^0$

 ${∆G_{rxn}}^0=\left(1 mol \left(-137.2\frac{kJ}{mol}\right)+3 mol \left(0.0\frac{kJ}{mol}\right)\right)- \left(1 mol \left(-50.8\frac{kJ}{mol}\right)+1 mol \times \left(-228.6\frac{kJ}{mol}\right)\right)$

${∆G_{rxn}}^0=142.2 kJ$

$C{H}_4 \left(g\right)+H_{2}O \left(g\right)\to CO \left(g\right)+3H_2 \left(g\right) ∆G_{{rxn}_1}°=142.2 kJ$

Calculate the standard free energy of the second reaction:

$CO \left(g\right)+H_{2}O \left(g\right)\to H_2 \left(g\right)+ C{O}_2 \left(g\right)$

${∆G_{rxn}}^0=\sum n_{product}∆{G_{f, product}}^0+ \sum n_{reactant}∆{G_{f, reactant}}^0$

 ${∆G_{rxn}}^0=\left(1 mol \left(0.0\frac{kJ}{mol}\right)+1 mol \left(-394.4\frac{kJ}{mol}\right)\right)- \left(1 mol \left(-137.2\frac{kJ}{mol}\right)+1 mol \times \left(-228.6\frac{kJ}{mol}\right)\right)$

${∆G_{rxn}}^0=-28.6 kJ$

$CO \left(g\right)+H_{2}O \left(g\right)\to H_2 \left(g\right)+ C{O}_2 \left(g\right) ∆G_{rxn}°=-28.6 kJ$

Now add the first and second reactions to get the final answer.

$C{H}_4 \left(g\right)+2H_{2}O \left(g\right)\to {CO}_2 \left(g\right)+4H_2 \left(g\right)$

${∆G_{overall}}^0=142.2 kJ+\left(-28.6 kJ\right)=113.6 kJ$

Overall $∆G°$ for the formation of $H_2+ C{O}_2$ from methane and steam is

$C{H}_4 \left(g\right)+2H_{2}O \left(g\right)\to {CO}_2 \left(g\right)+4H_2 \left(g\right) ∆G_{overall}°=113.6 kJ$

Conclusion:

We assigned oxidation numbers from oxidation number of O and H and calculated the standard free energy from the values of standard free energy of reaction given in the text.