Chapter 9, Problem 9.76QA

Interpretation Introduction

To calculate:

The $∆H^0$ of the reaction for the production of SNG.

Answer to Problem 9.76QA

Solution:

$∆H^0$ for the reaction $4C{O}_{\left(g\right)}+8 H_{2\left(g\right)} \to 3 C{H}_{4\left(g\right)}+C{O}_{2\left(g\right)}+2 H_2{O}_{\left(g\right)}$ is $-747.5 kJ.$

Explanation of Solution

1) Concept:

To calculate the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction, we can use Hess’s law. We can algebraically manipulate the four equations for which $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ values are known so that we can calculate the $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction. Then by applying Hess’s law, we can calculate the unknown value.

2) Given:

i) $C_{\left(graphite\right)}+2 H_{2\left(g\right)} \to {CH}_{4\left(g\right)} ∆H^0=-74.8 kJ$

ii) $C_{\left(graphite\right)}+\frac{1}{2}{O}_{2\left(g\right)} \to { CO}_{\left(g\right)} ∆H^0=-110.5 kJ$

iii) ${CO}_{\left(g\right)}+ \frac{1}{2}{O}_{2\left(g\right)}\to { CO}_{2\left(g\right) } ∆H^0=-283.0 kJ$

iv) $H_{2\left(g\right)}+ \frac{1}{2}{O}_{2\left(g\right)} \to H_2{O}_{\left(g\right)} ∆H^0=-285.8 kJ$

3) Calculations:

One of the major reactions in the production of SNG is shown as

$4C{O}_{\left(g\right)}+8 H_{2\left(g\right)} \to 3 C{H}_{4\left(g\right)}+C{O}_{2\left(g\right)}+2 H_2{O}_{\left(g\right)} ∆H^0= ?$

We combine equation A, B, C, and D to get the desired equation from the given reactions.

We have to multiply reaction A by 3 to get 3CH₄ in the final reaction.

A. $3 C_{\left(graphite\right)}+6 H_{2\left(g\right)} \to 3 {CH}_{4\left(g\right)} ∆H^0= -74.8\times 3=-224.4 kJ$

We have to reverse the reaction B and multiply by 3 to get  $3 C_{\left(graphite\right)}$ on product side so that we can cancel it in final reaction.

B.  ${3 CO}_{\left(g\right)}\to 3 C_{\left(graphite\right)}+\frac{3}{2}{O}_{2\left(g\right)} ∆H^0= +110.5 \times 3=+331.5 kJ$

We have to multiply reaction D by 2 to get 2H₂O in the final reaction.

D.  $2 H_{2\left(g\right)}+ O_{2\left(g\right)} \to {2 H}_2{O}_{\left(g\right)} ∆H^0= -285.8\times 2=-571.6 kJ$

Now add all four equations to get the final reaction:

$3 C_{\left(graphite\right)}+6 H_{2\left(g\right)} \to 3 {CH}_{4\left(g\right)} ∆H^0= -224.4 kJ$

$+{3 CO}_{\left(g\right)}\to 3 C_{\left(graphite\right)}+\frac{3}{2}{O}_{2\left(g\right)} ∆H^0=+331.5 kJ$

$+ {CO}_{\left(g\right)}+ \frac{1}{2}{O}_{2\left(g\right)}\to { CO}_{2\left(g\right) } ∆H^0= -283.0 kJ$

$+ 2 H_{2\left(g\right)}+ O_{2\left(g\right)} \to {2 H}_2{O}_{\left(g\right)} ∆H^0=-571.6 kJ$

$4C{O}_{\left(g\right)}+8 H_{2\left(g\right)} \to 3 C{H}_{4\left(g\right)}+C{O}_{2\left(g\right)}+2 H_2{O}_{\left(g\right)} ∆H^0=-747.5 kJ$

Conclusion:

We have calculated $∆H^0$ of the reaction by using Hess’s law and by manipulating the reactions.