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Consider the following elementary reaction:

2NO(g) → N2O2(g)

Suppose we let k1 stand for the rate constant for this reaction, and k-1 stand for the rate constant of the reverse reaction. Write the expression that gives the equilibrium concentration of NO in terms of k1, k-1. And the equilibrium concentration of N2O2.

Sagot :

Sebassandinidn

Answer:

[tex][NO]=\frac{k_{-1}}{k_1} [N_2O_2][/tex]

Explanation:

Hello!

In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:

[tex]r=-k_1[NO]+k_{-1}[N_2O_2][/tex]

However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:

[tex]0=-k_1[NO]+k_{-1}[N_2O_2][/tex]

Which can be also written as:

[tex]k_1[NO]=k_{-1}[N_2O_2][/tex]

Then, we solve for the concentration of NO to obtain:

[tex][NO]=\frac{k_{-1}}{k_1} [N_2O_2][/tex]

Best regards!

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