Under standard conditions, Delta G degree = -32.74 kJ/mol for the reaction N_2(g

Question

Under standard conditions, Delta G degree = -32.74 kJ/mol for the reaction N_2(g) + 3H_2(g) 2NH_3(g) What is the value of Delta G for the reaction at 298 K when the partial pressures of a mixture are 0.0100 atm for N_2(g), 0.0100 atm for H_2(g), and 5.00 atm for NH_3(g)? Is the reaction spontaneous or nonspontaneous in the forward direction under these conditions? In which direction must the reaction proceed to reach equilibrium? Calculate the equilibrium constant K for the above reaction at 25 degree C. Does K represent K_p or K_c?

Explanation / Answer

a) Kp = (Pnh3)^2/((Pn2)*(Ph2)^3)

= 5^2/((0.01^3)*(0.01))

=2.5*10^9

so G = dGo - RTln(k)

= -32.74 - 8.31*298*ln(2.5*10^9)*10^-3

=-86.4 kJ

b) SInce dG is negative the reaction is spontaneous under the given conditions

c) the reaction must proceed in the backward direction to attain equilibrium.

d) at equilibrium,

-dGo = -RTln(k)

or 32.74*10^3 = -8.73*298 *ln(k)

or K=1.4*10^15

the K represents the Kp since we are working with pressures.

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