Question 7.15 At 700 K, equilibrium constant for the reaction
H2(g) + I2(g) <---> 2HI(g)
is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Class XI Equilibrium Page 225
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H2(g) + I2(g) <=> 2HI(g) ; Kc = 54.8
when reaction will be reverse
2HI(g) <=> H2(g) + I2(g) , then, equilibrium constant of reverse reaction is K'c = 1/Kc = 1/54.8
A/C to question,
[HI] at equilibrium = 0.5 mol/L
Let at equilibrium conc. of [H2] = [I2] = x mol/L
K'c = [H2][I2]/[HI]²
1/54.8 = x.x/(0.5)²
1/54.8 = x²/(0.5 )²
x² = 0.25/54.8 = 0.00456
x = 0.0675 M
[H2] = [I2] = 0.0675 M
when reaction will be reverse
2HI(g) <=> H2(g) + I2(g) , then, equilibrium constant of reverse reaction is K'c = 1/Kc = 1/54.8
A/C to question,
[HI] at equilibrium = 0.5 mol/L
Let at equilibrium conc. of [H2] = [I2] = x mol/L
K'c = [H2][I2]/[HI]²
1/54.8 = x.x/(0.5)²
1/54.8 = x²/(0.5 )²
x² = 0.25/54.8 = 0.00456
x = 0.0675 M
[H2] = [I2] = 0.0675 M
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