Question

1. The equilibrium constant, Kp, for the following reaction is 4.55×10-5 at 723 K. N2(g) +...

1. The equilibrium constant, Kp, for the following reaction is 4.55×10-5 at 723 K.

N2(g) + 3H2(g) 2NH3(g)

If ΔH° for this reaction is -111 kJ, what is the value of Kp at 839 K?

Kp=

2. The equilibrium constant, Kp, for the following reaction is 0.110 at 298 K.
NH4HS(s) NH3(g) + H2S(g)


If ΔH° for this reaction is 92.7 kJ, what is the value of Kp at 393 K?

Kp=

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Homework Answers

Answer #1

1)

Given:

T1 = 723 K

T2 = 839 K

K1 = 4.55*10^-5

Ho = -111 KJ

= -111000 J

use:

ln(K2/K1) = (Ho/R)*(1/T1 - 1/T2)

ln(K2/4.55*10^-5) = (-111000/8.314)*(1/723 - 1/839.0)

ln(K2/4.55*10^-5) = -13351*(1.912*10^-4)

K2 = 3.54*10^-6

Answer: 3.54*10^-6

2)

Given:

T1 = 298 K

T2 = 393 K

K1 = 0.11

Ho = 92.7 KJ/mol

= 92700 J/mol

use:

ln(K2/K1) = (Ho/R)*(1/T1 - 1/T2)

ln(K2/0.11) = (92700.0/8.314)*(1/298 - 1/393.0)

ln(K2/0.11) = 11150*(8.112*10^-4)

K2 = 9.319*10^2

Answer: 9.32*10^2

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