Under standard conditions at 25 °C, the reaction ½ N 2 (g) + ½ O 2 (g) -> NO(g)

Question

Under standard conditions at 25 °C, the reaction ½ N2(g) + ½ O2(g) -> NO(g) has G° = +27.5 kJ/mol.

For the non-standard initial conditions, P(N2) = 1.0 bar, P(O2) = 1.0 bar, P(NO) = 10.0 bar at the same temperature, this reaction is:

Spontaneous in the forward direction, because G is a negative value

Spontaneous in the reverse direction, because K < 1

Spontaneous in the forward direction, because K > 1

Spontaneous in the reverse direction, because G is a positive value

Spontaneous in the forward direction, because G is a negative value

Spontaneous in the reverse direction, because K < 1

Spontaneous in the forward direction, because K > 1

Spontaneous in the reverse direction, because G is a positive value

Explanation / Answer

Calculate the equilibrium constant as below:

½ N2 (g) + ½ O2 (g) ------> NO (g); G0 = +27.5 kJ/mol.

Under non-standard conditions,

K = (PNO)/(PN2)1/2(PO2)1/2 = (10.0 bar)/(1.0 bar)1/2(1.0 bar)1/2 = (10.0 bar)/(1.0 bar) = 10.0

Therefore, G = G0 + RTln K where T = 25C = 298 K (standard temperature); plug in values and obtain

G = G0 + (8.314 J/mol.K)*(298 K)*ln (10.0) = (+27.5 kJ/mol) + (2.4776 kJ/mol)*ln (10.0) = (+27.5 kJ/mol) + (2.4776 kJ/mol)*(2.3026) = 30.2049 kJ/mol 30.20 kJ/mol.

Therefore, the forward reaction is non-spontaneous, since G has a high positive value; the forward reaction has a positive G which would indicate that the reverse reaction will have a negative G; therefore, the reverse reaction will be spontaneous. Hence, (d) is the correct option (ans).

Note that K > 1, but the value of K will hold no importance since G is positive and hence the forward reaction is non-spontaneous.

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