The $\Delta_{r}G$ of a reaction $A + B \rightleftharpoons C + D$ at fixed concentration of C is $-183.14 \, \text{kJ/mol}$. It is plotted as a function of temperature. Calculate the $G$ of the reaction (in kJ/mol).
Show Hint
For $\Delta G$ vs $T$ graph: slope = $-\Delta S$ and intercept = $\Delta H$.
Step 1: Set up the Gibbs-Helmholtz line as two unknowns.
\[ \Delta G = \Delta H - T \Delta S \]
Treat $\Delta H$ and $\Delta S$ as two unknowns to be found from two points read off the graph, rather than computing a slope first.
Step 2: Write one equation for each point on the graph.
At $T_1 = 200$ K, $\Delta G_1 = -183.14$ kJ/mol:
\[ \Delta H - 200 \Delta S = -183.14 \quad \text{...(i)} \]
At $T_2 = 300$ K, $\Delta G_2 = -100$ kJ/mol:
\[ \Delta H - 300 \Delta S = -100 \quad \text{...(ii)} \]
Step 3: Eliminate $\Delta H$ by subtracting the equations.
Subtracting (ii) from (i) removes $\Delta H$ since it appears with the same coefficient in both:
\[ (-200\Delta S) - (-300\Delta S) = -183.14 - (-100) \]
\[ 100 \Delta S = -83.14 \]
\[ \Delta S = -0.8314 \, \text{kJ mol}^{-1}K^{-1} \]
Step 4: Substitute back into equation (i) to get $\Delta H$.
\[ \Delta H = -183.14 + 200(-0.8314) \]
\[ \Delta H = -183.14 - 166.28 = -349.42 \, \text{kJ/mol} \]
Step 5: Round to match the answer choices.
\[ \Delta H \approx -349 \, \text{kJ/mol} \]
Step 6: This value is what the graph's vertical intercept represents.
Since $\Delta H$ is the value $\Delta G$ would take at $T=0$, it is exactly what the question calls the "G of the reaction" here.
