Question

for the reaction 2Hcl -----H2 +cl2 the kc is 1.0×10^-5 if the eqlibrium conc of H2 nd cl2 are 1.2×10-3 nd 1.2×10^-4 respecitly conc of hcl is?

Answer 1

Let initial number of moles of HCl be N in the container of volume V litres.  Let there be X moles of H2 and X moles of Cl2 at the equilibrium.

       2 H Cl (g)  <===> H2 (g)   +   Cl2 (g)
      N-2X moles       X moles    X moles 

Equilibrium constant Kc = [ H2] [ Cl2] / [HCl ]²
      Kc = (X/V)² ÷ {(N-2X) / V}² = 4 * 10⁻³⁴ 
           = X² / (N-2X)²  = 4 * 10⁻³⁴

This means that  X/(N-2X ) = 2 * 10⁻¹⁷
          N-2X = 5 * 10¹⁶ X
          X ≈ 2 * 10⁻¹⁷  N

So X is very very small compared to N.. Equilibrium shifts towards the left.

Answer 2

Answer:

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