Question

At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 0.0033% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction.

a) 2.08 × 10^−5
b) 4.16 × 10^−5
c) 1.04 × 10^−4
d) 2.08 × 10^−4

Answer 1

The equilibrium constant for the decomposition reaction of NO2 into NO and O2 at the given conditions is 4.16 × 10−5, corresponding to option (b).The correct option is b.

Step-by-step explanation:

The student's question is asking to calculate the equilibrium constant (K) for a decomposition reaction of NO2 into NO and O2 at certain conditions, given the percentage decomposition. In order to determine the answer, we must use the equilibrium expression that relates the concentrations of the products and reactants at equilibrium. For the reaction:

2NO2(g) → 2NO(g) + O2(g)

The K expression is:

K = [NO]2[O2]/[NO2]2

If NO2 is 0.0033% decomposed at equilibrium, this means that [NO2] has decreased by 0.0033% of 1.00 M, which is 0.000033 M. Therefore, 0.000033 M of NO2 has decomposed into 0.000033 M of NO and 0.0000165 M of O2 (half due to the stoichiometry). Substituting these values into the K expression:

K = (0.0000332 × 0.0000165) / (1.00 - 0.000033)2

Upon calculation, K will equal 4.16 × 10−5, which corresponds to option (b).

The correct option is b