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Question

A closed system containing 102M H2 and 2×102M I2 at 450oC is allowed to reach equilibrium and at equilibrium the HI concentration is 1.8×102M. Calculate KC at 450oC for H2(g)+I2(g)2HI(g)

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Solution

H2+I22HIInitial1022×1020Equilibrium102(2×102)x2x
given, at equilibrium [HI]=1.8×102M
kc=(2x)2(0.02x)(0.01x)=4x2x2+0.0002x(0.03)
=0.0003240.000081+0.00020.00027
equilibrium constant kc=29.455

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