Question

For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $\Delta_r G^\circ$ for the reaction at 300 K and 1 atm in J mol$^{-1}$ is $-xR$, where $x$ is __________. (Rounded off to the nearest integer) [R=8.31 J mol$^{-1}K^{-1}$ and $\ln 10=2.3$]

Hint:

If $K_{eq}>1$, $\Delta G^\circ$ is negative (spontaneous under standard conditions). If $K_{eq}<1$, $\Delta G^\circ$ is positive.

Answer 1

Correct Answer: 1380

Step 1: Relation: $\Delta_r G^\circ = -RT \ln K_{eq}$.
Step 2: Substitute values: $\Delta_r G^\circ = -R \times 300 \times \ln(100)$.
Step 3: Since $100 = 10^2$, $\ln(100) = 2 \ln(10)$.
Step 4: Given $\ln 10 = 2.3$, so $\ln(100) = 2 \times 2.3 = 4.6$.
Step 5: $\Delta_r G^\circ = -300 \times R \times 4.6 = -1380 R$.
Step 6: Comparing with $-xR$, we get $x = 1380$.