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What wI'll be equilibrium constant at 127°c. If equilibrium at 27°c is 4 for reaction. N2 +3H2 = 2NH3; ​^H = -46.06KJ.

What wI'll be equilibrium constant at 127°c. If equilibrium at 27°c is 4 for reaction. N2 +3H2 = 2NH3;  ​^H = -46.06KJ.
 

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1 Answers

Arun
25750 Points
5 years ago
Dear Faisal
 
Use vont Hoff equation.
 
log (k2 /k1) = 2.303 /R (1/T1 – 1/T2)
 

K2- Equilibrium constant at 27C.

K1-Equilibrium constant at 127C.

R- Gas constant(0.008314 for answer in KJ)

Change in enthalpy.

T2- 400K(127C)

T1- 300K(27C).

Solve correctly and get the answer.

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