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At 850K, the value of the equilibrium constant Kp for the ammonia synthesis reaction: N2(g)+H2(g)--> N2H2(g)...

At 850K, the value of the equilibrium constant Kp for the ammonia synthesis reaction: N2(g)+H2(g)--> N2H2(g) is 0.3290. If a vessel contains an initial reaction mixture in which [N2]= 0.0200 M, [H2]=0.0200 M, and [N2H2]= 0.000150 M, what will the [N2H2] be when equilibrium is reached? The answer is not: 0.000298, .00681, .000172, .000148, .00031, .00471 please help!

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Expert Solution


initially

N2(g)+H2(g) <--> N2H2(g)

reaction quotient(Q) = [N2H2]/[N2][H2]

     QC = 0.00015/(0.02*0.02) = 0.375

KP = KC(RT)^Dn

Dn = 1-2 = -1

R = 0.0821 l.atm.k-1.mol-1

T = 850 k

Kp = 0.329

0.329 = Kc*(0.0821*850)^-1

on solving

Kc = 22.96

Qc < Kc . so that forward reaction takes place

            N2(g) + H2(g) <--> N2H2(g)

initial     0.02 M   0.02 M     0.00015 M

change       x         x           x

equilibrium 0.02-x   0.02-x      0.00015+x

   KC = [N2H2]/[N2][H2]

   22.96 = (0.00015+X)/(0.02-X)^2

X = 0.005 M

so that,

at equilibrium

[N2H2] = 0.00015+x = 0.005+0.00015 = 0.00515 M

queen_honey_blossom answered 2 years ago
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