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For the given EQ. L(gas) ---------> M(gas), K(forward) = 5 * 10^-4 mol/L/Sec K(back) = 3 * 10^-2 L/Mol/sec, then equilibrium concentration of M is

For the given EQ.
L(gas) ---------> M(gas), 
K(forward) = 5 * 10^-4 mol/L/Sec
K(back) = 3 * 10^-2 L/Mol/sec, then
equilibrium concentration of M is

Grade:12th pass

1 Answers

Ramreddy IIT Patna
askIITians Faculty 49 Points
9 years ago
We know that KC= kf/kb = [M]/[L]
since Delta n= 0 here Kp=Kc
Kc= 5 x 10-4 / 3 x 10 -2
Kc = 1.66 x 10-2
[M]/[L] = 1.66 x 10-2
L(gas) ---------> M(gas)
Assume that intialmoles of L is 1. Then
1 0
1-x x (At equilibrium)
Kc= x/1-x = 1.66 x 10-2
After solving we will get x= 1.66/2.66
= 0.624
Equilibriumconcentration of M is = 0.624/V (V is volume of reaction mixture).

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