The equilibrium constant for the reaction given below is 2.0×10−7 at 300K. Calculate the standard entropy change for the reaction. (ΔH0=28.4kJmol−1) PCl5(g)⇌PCl3(g)+Cl2(g)
−3.36Jmol−1
−33.6Jmol−1
−0.336Jmol−1
−336Jmol−1
A
−0.336Jmol−1
B
−33.6Jmol−1
C
−336Jmol−1
D
−3.36Jmol−1
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Solution
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Given,
T=300KΔH=28.4 KJ/mol
K=2×10−7
We have,
ΔG0=−RTlnK
=−8.314×300×ln(2×10−7)
=−38472.90J
ΔG0=−38.47KJ/mol
and now, we have
ΔG0=ΔH−TΔS
−38.47=28.4−300×ΔS
ΔS=−38.47−28.4−300
ΔS=0.33KJmolK
Ans :- Option B.
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