The equilibrium constant (K) of the reaction below is K=6.0*10^-2, with initial
ID: 813333 • Letter: T
Question
The equilibrium constant (K) of the reaction below is K=6.0*10^-2, with initial concentrations as follows: [H2]=1.0*10^-2 M, [N2]=4.0 M, and [NH2]=1.0*10^-4 M.
N2 (g) + 3H2 (g)--><-- 2NH3 (g)
1. If the concentration of the reactant H2 was increased from 1.0*10^-2 M to 2.5 *10^-2 M, calculate the reaction quotient (Q) and determine which way the equilibrium position would shift.
2. If the concentration of the reactant H2 was decreased from 1.0*10^-2 M to 2.7*10^-4 M, calculate the reaction quotient (Q) and determine which way the equilibrium position would shift.
3. If the concentration of the product NH3 was decreased from 1.0*10^-4 M to 5.6 *10^-3 M, calculate the reaction quotient (Q) and determine which way the equilibrium position would shift.
PLEASE SHOW ALL WORK
Explanation / Answer
1) Qc= [NH3]^2/[N2][H2]^3
= (1.0*10^-4)^2/ [2.5X10^-2]^3 X4
= 0.016 X10^-2
K=6.0*10^-2 Qc < Kc
reaction proceeds to left to right
2)Qc= (1.0*10^-4)^2/ [2.7X10^-4]^3 X4
= 127
Qc> Kc
reaction proceeds to right to left
3) Qc= (5.6*10^-3)^2/ [1X10^-2]^3 X4
= 7.84
Qc> Kc
reaction proceeds to right to left
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