Question

In the reaction \( \frac{3}{2}O_2(g) \rightarrow O_3(g) \), the value of \( \Delta_r G^\circ \) at \( 298\,K \) is approximately ( \( K_p = 10^{-30},\ 2.303RT = 5.7\,\text{kJ mol}^{-1} \) )

• A. \(171\,\text{kJ mol}^{-1} \)
• B. \(191\,\text{kJ mol}^{-1} \)
• C. \(-171\,\text{kJ mol}^{-1} \)
• D. \(-191\,\text{kJ mol}^{-1} \)
• E. \(100\,\text{kJ mol}^{-1} \)

Hint:

\(\log 10^{-n} = -n\)

Answer 1

The Correct Option is A

Step 1: Understanding Gibbs Free Energy and Equilibrium Constant:
The standard Gibbs free energy change (\(\Delta_r G^\ominus\)) of a reaction is related to its equilibrium constant (\(K_p\) for gas-phase reactions). A very small equilibrium constant (\(K_p \ll 1\)) indicates that the reaction is non-spontaneous in the forward direction, which corresponds to a large positive \(\Delta_r G^\ominus\).
Step 2: Key Formula or Approach:
The relationship between \(\Delta_r G^\ominus\) and \(K_p\) is given by the equation:
\[ \Delta_r G^\ominus = -RT \ln K_p \] This can be written in terms of the base-10 logarithm as:
\[ \Delta_r G^\ominus = -2.303 RT \log_{10} K_p \] Step 3: Detailed Calculation:
We are given the following values:
\(K_p = 10^{-30}\)
\(2.303 RT = 5.7\) kJ mol\(^{-1}\)
Substitute these values into the formula:
\[ \Delta_r G^\ominus = -(5.7 \text{ kJ mol}^{-1}) \times \log_{10}(10^{-30}) \] Using the logarithm property \(\log(a^b) = b \log(a)\):
\[ \log_{10}(10^{-30}) = -30 \times \log_{10}(10) = -30 \times 1 = -30 \] Now, substitute this back into the equation:
\[ \Delta_r G^\ominus = -(5.7 \text{ kJ mol}^{-1}) \times (-30) \] \[ \Delta_r G^\ominus = 5.7 \times 30 \text{ kJ mol}^{-1} \] \[ \Delta_r G^\ominus = 171 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The value of \(\Delta_r G^\ominus\) is 171 kJ mol\(^{-1}\). The positive value is consistent with the very small \(K_p\), indicating the formation of ozone from oxygen under standard conditions is highly non-spontaneous.