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Interpretation:
The equilibrium constant of the given reaction have to be determined.
Concept Introduction:
aA+bB⇌cC+dD
Rate of the forward reaction rf = kf [A]a [B]b
Rate of the backward reaction rb = kb [C]c [D]d
kf and kb are the rate constants of forward and backward reaction respectively
At equilibrium the rate of the forward reaction = the rate of the backward reaction
kf [A]a [B]b = kb [C]c [D]d
kfkb = Keq = [C]c [D]d[A]a [B]b
Keq is the equilibrium constant of the reaction.
Answer to Problem 71A
The equilibrium constant of the given reaction is 1.0.
Explanation of Solution
Data given: At initially [N2] = 5.0 M and [H2] = 4.0 M and at equilibrium [H2] = 1.0 M
The given chemical equation is-
N2(g)+3H2(g)⇌2NH3(g)
When t = 0, [N2] = 5.0 M and [H2] = 4.0 M
At equilibrium, [N2] = (5.0 - x) M, [H2] = (4.0 - 3x) M and [NH3] = 2x M
Given that, [H2] = (4.0 - 3x) M = 1.0 M
So, the value of x = 1
At equilibrium, [N2] = 4.0 M, [H2] = 1.0 M and [NH3] = 2.0 M
Keq = [NH3]2[N2] [H2]3
= (2.0)2(4.0) × (1.0)3 = 1.0
Chapter 17 Solutions
World of Chemistry, 3rd edition
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