The equilibrium constant (kc) for the reaction 2HCl(g) =H2(g)+Cl2 (g) is 4×10^-34 at 25°C. What is equilibrium constant for the reaction?
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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.
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.
kvnmurty:
:-)
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