Q.
Initially 0.8 mole of PCl5 and 0.2 mole of PCl3 are mixed in one litre vessel. At equilibrium 0.4 mole of PCl3 is present. The value of Kc for the reactionPCl5(g)⇔PCl3(g)+Cl2(g)
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a
0.13molL−1
b
0.5molL−1
c
0.065molL−1
d
0.1molL−1
answer is A.
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Detailed Solution
PCl5⇌PCl3+Cl2 Given (0.2+x)=0.4 0.8 0.2 0 ⇒x=0.2 x x x_ (0.8−x) 0.2+x x ∴Kc=0.4×0.20.6 =0.133 M =0.133 mole.lit−1
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