A reaction has an equilibrium constant of Kp=0.025 at 27 C. Find Grxn for the re

Question

A reaction has an equilibrium constant of Kp=0.025 at 27 C. Find Grxn for the reaction at this temperature.

-9.20 kJ

Determine G for the following reaction:

2NO(g)+O2(g)N2O4(g)

Use the following reactions with known Grxn values:

N2O4(g)2NO2(g), Grxn = 2.8 kJ
NO(g)+12O2(g)NO2(g), Grxn = - 36.3 kJ

Express your answer using one decimal place.

-9.20 kJ

Determine G for the following reaction:

2NO(g)+O2(g)N2O4(g)

Use the following reactions with known Grxn values:

N2O4(g)2NO2(g), Grxn = 2.8 kJ
NO(g)+12O2(g)NO2(g), Grxn = - 36.3 kJ

Express your answer using one decimal place.

Explanation / Answer

1)

we know that

for a reaction

dGo = -RTlnKp

given

Kp = 0.025

R = 8.314

temperature (T) = 27 C = 27 + 273 = 300 K

so

using those values

dGo = -8.314 x 300 x ln 0.025

dGo = -9200

dGo = -9.2 x 1000 J

dGo = -9.2 kJ

so

the value of dGo for the reaction is -9.2 kJ

2)

consider the given reactions

a) N204 ----> 2 N02 ----> dGoa = 2.8

b) N0 + 0.5 02 ---> N02 -----> dGob = -36.3

c) 2 N0 + 02 ---> N204 ----> dGoc = ??

we can see that

eq c = ( 2 x eqb) - ( eq a)

so

we get

dGoc = (2 x dGob) - (dGoa)

using given values

dGo c = ( 2 x -36.3) - (2.8)

dGo c = -75.4 kJ

so

the value of dGo for the required reaction is -75.4 kJ

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