Question:
For a certain reaction, \(\Delta \mathrm{H}^{\circ}=219 \mathrm{~kJ}\) and \(\Delta \mathrm{S}^{\circ}=-21 \mathrm{JK}^{-1}\). Find the value of \(\Delta \mathrm{G}^{\circ}\).
For a certain reaction, \(\Delta \mathrm{H}^{\circ}=219 \mathrm{~kJ}\) and \(\Delta \mathrm{S}^{\circ}=-21 \mathrm{JK}^{-1}\). Find the value of \(\Delta \mathrm{G}^{\circ}\).
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Use \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) at the standard temperature of 298 K, converting entropy to kJ.
Updated On: Nov 13, 2026
202.74 kJ
-312.74 kJ
225.26 kJ
-325.06 kJ
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
\(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) with \(T=298\ \text{K}\) (standard).
Step 2: Unit conversion
\(\Delta S^{\circ}=-21\ \mathrm{J\,K^{-1}}=-0.021\ \mathrm{kJ\,K^{-1}}\).
Step 3: Substitute
\(\Delta G^{\circ}=219-298\times(-0.021)=219+6.258\).
Step 4: Result
\(\Delta G^{\circ}\approx225.26\ \mathrm{kJ}\).
Final Answer:
The standard Gibbs energy change is about 225.26 kJ. \[ \boxed{\text{(C) }225.26\text{ kJ}} \]
\(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) with \(T=298\ \text{K}\) (standard).
Step 2: Unit conversion
\(\Delta S^{\circ}=-21\ \mathrm{J\,K^{-1}}=-0.021\ \mathrm{kJ\,K^{-1}}\).
Step 3: Substitute
\(\Delta G^{\circ}=219-298\times(-0.021)=219+6.258\).
Step 4: Result
\(\Delta G^{\circ}\approx225.26\ \mathrm{kJ}\).
Final Answer:
The standard Gibbs energy change is about 225.26 kJ. \[ \boxed{\text{(C) }225.26\text{ kJ}} \]
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