Question

Consider the reaction
$\mathbf{2}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$
a. Using data from Appendix 4, calculate K at 298 K.
b. What is for this reaction at T = 298 K if the reactants, each at 10.0 atm, are changed to products at 10.0 atm? (Hint: Construct a thermodynamic cycle and consider how entropy depends on pressure.)

Short answer

1. The value of Kat 298 K is $1.24\times {10}^{20}$.
2. The value of $∆\mathrm{S}$ for this reaction at T = 298 K if the reactants, each at 10.0 atm, are changed to products at 10.0 atm is -154K/J .

Step-by-step solution

 Introduction to the concept

The following is the relationship between Gibbs free energy and equilibrium constant;

$∆\mathrm{G}°=-\mathrm{RTlnK}$

Where

R= Universal gas constant,

T=temperature, and

K=equilibrium constant.

Determination of equilibrium constant K

Here is the given reaction

$2\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

The change in standard Gibbs free energy for the aforementioned reaction can be calculated as follows:

$$\begin{gathered} ∆\mathrm{G}°=2∆\mathrm{G}°{\mathrm{CO}}_2\left(\mathrm{g}\right)-2∆\mathrm{G}°\mathrm{CO}\left(\mathrm{g}\right)-∆\mathrm{G}°\left(\mathrm{g}\right) \\ ∆\mathrm{G}°=-514\raisebox{1ex}{$kJ$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \end{gathered}$$

The following is the relationship between Gibbs free energy and equilibrium constant:

$∆\mathrm{G}°=-\mathrm{RTlnK}$

In terms of putting the numbers together,

$$\begin{gathered} \mathrm{K}=\mathrm{e} \\ =1.24\times {10}^{20} \end{gathered}$$

After rearranging

As a result, the value of the equilibrium constant is $1.24\times {10}^{20}$

Determination of equilibrium constant K

b)

At 1.00 atm, the conventional entropy values are used. For 2 mol of carbon monoxide gas, the change in entropy may be estimated by changing the pressure from 10 atm to 1 atm.

$∆\mathrm{S}=\mathrm{nRln}\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

By substituting the values,

$∆\mathrm{S}=38.3\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

Likewise, for a change in temperature of 1 mol of oxygen gas from 10 atm to 1 atm, the change in entropy can be calculated as follows:

$∆\mathrm{S}=\mathrm{nRln}\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

By substituting the values,

$∆\mathrm{S}=19.1\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

The change in entropy of a reaction may now be estimated using a balanced chemical equation and reactant and product quantities as follows

The entropy required to transform 2 mol of carbon dioxide from 1 atm to 10 atm can be calculated as follows:

$∆\mathrm{S}=-38.3\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

The following formula can be used to compute total entropy:

$$\begin{gathered} ∆\mathrm{S}=38.3+19.1-173-38.3 \\ =-154\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \end{gathered}$$

As a result, the reaction's total entropy is .$-154\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$