At 700 K, the equilibrium constant for the reaction NO2(g) « 2NO(g) + O2(g) is Kc = 4.79×10-3 (M), and the rate constant for the reaction 2NO(g) + O2(g) ® NO2(g) is k = 3.13×10^3 M^-2s^-1.

Question

At 700 K, the equilibrium constant for the reaction NO2(g) « 2NO(g) + O2(g) is Kc = 4.79×10-3 (M), and the rate constant for the reaction 2NO(g) + O2(g) ® NO2(g) is k = 3.13×10^3 M^-2s^-1.

What is the rate constant for the reaction NO2(g) ® 2NO(g) + O2(g) at this temperature?

Answer 1

it will be 75.5 dgree celcuos

Answer 2

To determine the rate constant for the reaction NO2(g) → 2NO(g) + O2(g) at a certain temperature, we can use the concept of the equilibrium constant.

The equilibrium constant (Kc) for a reaction relates the concentrations of reactants and products at equilibrium. In this case, the equilibrium constant is given as Kc = 4.79 x 10^(-3) (M).

The equilibrium constant can be obtained by dividing the concentration of the products by the concentration of the reactants, each raised to the power of their respective stoichiometric coefficients.

For the given reaction:
Kc = [NO]^2 [O2] / [NO2]
Since [NO] and [O2] are products and [NO2] is a reactant, we can rewrite the equilibrium constant expression as:
Kc = [NO]^2 [O2] / [NO2]^1

To find the rate constant for the reverse reaction, we can use the concept of the equilibrium constant and the rate constant for the forward reaction.

The rate of a chemical reaction is determined by the rate constant (k) multiplied by the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients. In this case, the rate constant for the forward reaction is given as k = 3.13 x 10^3 M^(-2) s^(-1).

For the forward reaction:
rate = k[NO]^2 [O2]

For the reverse reaction, we can consider it as the forward reaction in reverse:
rate = k_reverse[NO2]

Since the rate of the reverse reaction is equal to the rate of the forward reaction, we can equate the rate expressions and solve for the rate constant of the reverse reaction.

k_reverse[NO2] = k[NO]^2 [O2]

Rearranging the equation, we find:
k_reverse = (k[NO]^2 [O2]) / [NO2]

Now, plug in the given values:
k_reverse = (3.13 x 10^3 M^(-2) s^(-1) * ([NO]^2 [O2])) / [NO2]

Remember that the concentrations of [NO] and [O2] are unknown and not provided in the question. To determine them, you need additional information such as the initial concentrations or a rate law expression involving the concentrations of NO and O2.

Without this additional information, we can't calculate the rate constant for the reverse reaction.