Question

Consider the following reaction at 298 K: N2O4(g)⇌2NO2(g) with KP=0.142. Under what conditions will the reaction proceed spontaneously?

a) When the concentration of N2O4 is high and the concentration of NO2 is low.
b) When the concentration of NO2 is high and the concentration of N2O4 is low.
c) The reaction is not spontaneous under any conditions.
d) Spontaneity depends on the initial concentrations.

Answer 1

From the given value of Kp = 0.142, we can infer that the reaction is non-spontaneous under the given conditions, as Kp is less than 1. Therefore, the correct answer is:

• c) The reaction is not spontaneous under any conditions.

Step-by-step explanation:

The spontaneity of a reaction is determined by the sign of the Gibbs free energy change (ΔG). If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. The relationship between ΔG and the equilibrium constant (Kp) is given by the equation ΔrG° = -RT ln(Kp), where ΔrG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and ln(Kp) is the natural logarithm of the equilibrium constant.

From the given value of Kp = 0.142, we can infer that the reaction is non-spontaneous under the given conditions, as Kp is less than 1. Therefore, the correct answer is: c) The reaction is not spontaneous under any conditions.