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Consider the following reaction:
[tex]\[ CO_2(g) + CCl_4(g) \rightleftharpoons 2 COCl_2(g) \][/tex]

Calculate [tex]\(\Delta G\)[/tex] for this reaction at [tex]\(25^{\circ}C\)[/tex] under these conditions:
[tex]\[
\begin{array}{ll}
P_{CO_2} & = 0.140 \text{ atm} \\
P_{CCl_4} & = 0.185 \text{ atm} \\
P_{COCl_2} & = 0.735 \text{ atm}
\end{array}
\][/tex]
[tex]\(\Delta G_f^{\circ}\)[/tex] for [tex]\(CO_2(g)\)[/tex] is [tex]\(-394.4 \text{ kJ/mol}\)[/tex], [tex]\(\Delta G_f^{\circ}\)[/tex] for [tex]\(CCl_4(g)\)[/tex] is [tex]\(-62.3 \text{ kJ/mol}\)[/tex], and [tex]\(\Delta G_f^{\circ}\)[/tex] for [tex]\(COCl_2(g)\)[/tex] is [tex]\(-204.9 \text{ kJ/mol}\)[/tex].

Express the energy change in kilojoules per mole to three significant figures.

[tex]\[
\Delta G = \square \text{ kJ/mol}
\][/tex]

Sagot :

GinnyAnsweridn
To calculate the Gibbs free energy change [tex]\(\Delta G\)[/tex] for the given reaction at 25°C under the specified conditions, we follow a systematic approach by breaking down the steps in the process:

1. Write the given reaction and specify the partial pressures:

[tex]\[ \text{CO}_2(g) + \text{CCl}_4(g) \rightleftharpoons 2 \text{COCl}_2(g) \][/tex]

- [tex]\( P_{\text{CO}_2} = 0.140 \, \text{atm} \)[/tex]
- [tex]\( P_{\text{CCl}_4} = 0.185 \, \text{atm} \)[/tex]
- [tex]\( P_{\text{COCl}_2} = 0.735 \, \text{atm} \)[/tex]

2. Identify the standard Gibbs free energy of formation ([tex]\(\Delta G_f^\circ\)[/tex]) for each substance:

- [tex]\(\Delta G_f^\circ (\text{CO}_2) = -394.4 \, \text{kJ/mol} \)[/tex]
- [tex]\(\Delta G_f^\circ (\text{CCl}_4) = -62.3 \, \text{kJ/mol} \)[/tex]
- [tex]\(\Delta G_f^\circ (\text{COCl}_2) = -204.9 \, \text{kJ/mol} \)[/tex]

3. Calculate the standard Gibbs free energy change ([tex]\(\Delta G^\circ\)[/tex]) of the reaction:

[tex]\[ \Delta G^\circ = \left( 2 \times \Delta G_f^\circ (\text{COCl}_2) \right) - \left( \Delta G_f^\circ (\text{CO}_2) + \Delta G_f^\circ (\text{CCl}_4) \right) \][/tex]

Substituting the values:

[tex]\[ \Delta G^\circ = \left( 2 \times -204.9 \, \text{kJ/mol} \right) - \left( -394.4 \, \text{kJ/mol} + (-62.3 \, \text{kJ/mol}) \right) = -409.8 \, \text{kJ/mol} - (-456.7 \, \text{kJ/mol}) = 46.9 \, \text{kJ/mol} \][/tex]

4. Calculate the reaction quotient (Q) at the given conditions:

The reaction quotient [tex]\( Q \)[/tex] is given by:

[tex]\[ Q = \frac{(P_{\text{COCl}_2})^2}{P_{\text{CO}_2} \cdot P_{\text{CCl}_4}} \][/tex]

Substituting the partial pressures:

[tex]\[ Q = \frac{(0.735)^2}{0.140 \times 0.185} = 20.8581 \][/tex]

5. Calculate the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) at the given conditions:

[tex]\[ \Delta G = \Delta G^\circ + RT \ln Q \][/tex]

Where:
- [tex]\( R \)[/tex] (Universal gas constant) = 0.008314 kJ/(mol K)
- [tex]\( T \)[/tex] (temperature in Kelvin) = [tex]\( 25^\circ \text{C} + 273.15 = 298.15 \, \text{K} \)[/tex]
- [tex]\( \ln Q \)[/tex] is the natural logarithm of Q

Substituting in the values:

[tex]\[ \Delta G = 46.9 \, \text{kJ/mol} + (0.008314 \, \text{kJ/(mol K)} \times 298.15 \, \text{K} \times \ln 20.8581) \][/tex]

[tex]\[ \Delta G = 46.9 \, \text{kJ/mol} + (0.008314 \times 298.15 \times 3.036) = 54.43 \, \text{kJ/mol} \][/tex]

Therefore, the Gibbs free energy change [tex]\(\Delta G\)[/tex] for the reaction at 25°C under the given conditions is 54.4 kJ/mol (rounded to three significant figures).
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