What is the value of the Gibbs energy change, \( \Delta G \), in \( \mathrm{kJ\,mol^{-1}} \), for the reaction between hydrogen and iodine gases at \(30^\circ\mathrm{C}\)?
\( \mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)} \qquad K=40.2 \)
\( \Delta G=-RT\ln K \qquad R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}} \)
Correct Answer: \( \boxed{\mathrm{C}} \)
Use the relationship:
\( \Delta G=-RT\ln K \)
Convert the temperature to kelvin:
\(T=30+273=303\,\mathrm{K}\)
Substitute the values:
\( \Delta G=-(8.31)(303)\ln(40.2) \)
\( \Delta G\approx-9300\,\mathrm{J\,mol^{-1}} \)
Convert to \( \mathrm{kJ\,mol^{-1}} \):
\( \Delta G\approx-9.30\,\mathrm{kJ\,mol^{-1}} \)
Therefore, the correct answer is (C).
Which reaction is predicted to occur with the largest decrease in entropy?
Correct Answer: \( \boxed{\mathrm{C}} \)
Entropy is a measure of the disorder or randomness of a system. Gases have much greater entropy than solids and liquids.
For reaction (C), the number of moles of gas decreases from:
\(6+6=12\,\mathrm{mol}\)
to:
\(6\,\mathrm{mol}\)
In addition, \( \mathrm{C_6H_{12}O_6} \) is formed as a solid. Thus, a large number of gaseous molecules are converted into fewer gas molecules together with a solid, producing a very large decrease in entropy.
The general relationship is:
\( \Delta S^\circ=\sum S^\circ_{\mathrm{products}}-\sum S^\circ_{\mathrm{reactants}} \)
Therefore, the correct answer is (C).
Entropy decreases when the system becomes more ordered.
(D) \( 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{g}) \): 3 moles of gas → 2 moles of gas, fewer gas particles means lower entropy.
✅ Answer: (D)
