The equilibrium constant for the reaction given below is 2.0×10−7 at 300 K. Calculate the standard free energy change for the reaction; PCl5(g)⇌PCl3(g)+Cl2(g) Also, calculate the standard entropy change if △Ho=28.40kJmol−1.
A
△Go=40kJmol−1△So=30JK−1mol−1
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B
△Go=38.48kJmol−1△So=−33.6JK−1mol−1
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C
△Go=45kJmol−1△So=−35JK−1mol−1
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D
△Go=42kJmol−1△So=40JK−1mol−1
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Solution
The correct option is B△Go=38.48kJmol−1△So=−33.6JK−1mol−1 The formula which relate Gibbs free energy ΔG0 and equilibrium constant Keq is ΔG0=−2.303×R×T×log10[Keq] △Go=−2.303×8.314×300log10[2.0×10−7] =+38479.8Jmol−1 =+38.48kJmol−1 Also, △Go=△Ho−T△So ∴△So=△Ho−△GoT △So=28.40−38.48300 △So=−0.0336kJK−1mol−1=−33.6JK−1mol−1