Question:
For a reaction, \(\Delta \text{G}^{\circ }=-5.7\) kJ mol\({}^{-1}\) at 298 K. Calculate K.
( R = 8.314 J K\({}^{-1}\) mol\({}^{-1}\) )
For a reaction, \(\Delta \text{G}^{\circ }=-5.7\) kJ mol\({}^{-1}\) at 298 K. Calculate K.
( R = 8.314 J K\({}^{-1}\) mol\({}^{-1}\) )
( R = 8.314 J K\({}^{-1}\) mol\({}^{-1}\) )
Show Hint
Use delta G standard equals minus 2.303 RT log K, with delta G converted to joules.
Updated On: Nov 12, 2026
- \(1\)
- \(10\)
- \(10^{2}\)
- \(10^{3}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
\(\Delta G^{\circ}=-2.303\,RT\log K\).
Step 2: Substitute
\(-5700=-2.303\times8.314\times298\times\log K\)
\(2.303\times8.314\times298\approx5705\).
Step 3: Calculation
\(\log K=\dfrac{5700}{5705}\approx1\), so \(K\approx10\).
Final Answer:
The equilibrium constant is about 10. \[ \boxed{\text{(B) }10} \]
\(\Delta G^{\circ}=-2.303\,RT\log K\).
Step 2: Substitute
\(-5700=-2.303\times8.314\times298\times\log K\)
\(2.303\times8.314\times298\approx5705\).
Step 3: Calculation
\(\log K=\dfrac{5700}{5705}\approx1\), so \(K\approx10\).
Final Answer:
The equilibrium constant is about 10. \[ \boxed{\text{(B) }10} \]
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