Consider: 2 NH3(g) 3 H2(g) + N2(g) The enthalpy change at 298 K for this reactio

Question

Consider:

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

The enthalpy change at 298 K for this reaction is Hrxn = 92.38 kJ/mol. Under the same conditions the entropy change for the reaction is Sorxn = 198.24 J/mol-K. Calculate the value of Keq for this reaction at 521 K. Assume that Hrxn and Srxn are the same at 521 K as they are at 298 K. Use R = 8.314 J/K for the value of the gas constant.


a) At this temperature Keq = 1.00e+00.
b) At this temperature Keq = 1.24e+01.
c) At this temperature Keq = 1.45e-06.
d) At this temperature Keq = 8.07e-02.
e) At this temperature Keq = 4.58e-04.

Explanation / Answer

Keq = [NO]^2 [Cl2] / [NClO]^2 = = ( 6.4)^2 x 0.49 / (1.6)^2 =7.84

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