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onsider the following elementary reaction:
NO_(g) +O_2(g) → NO_2(g) + O_(g)
Suppose we let k_1 stand for the rate constant of this reaction, and k_-1 stand for the rate constant of the reverse reaction. Write an expression that gives the equilibrium concentration of O_2 in terms of k_1 k_-1. and the equilibrium concentrations of NO NO_2 and O. [O_2] =

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sebassandin

Answer:

[tex]C_{O_2}=\frac{k_2C_{NO_2}C_{O}}{k_1C_{NO}}[/tex]

Explanation:

Hello,

In this case, for the given elementary reaction, the rate law takes the form:

[tex]r=-k_1C_{NO}C_{O_2}+k_2C_{NO_2}C_{O}[/tex]

Nevertheless, at equilibrium, the rate becomes zero:

[tex]0=-k_1C_{NO}C_{O_2}+k_2C_{NO_2}C_{O}[/tex]

Thus, we can solve for the concentration of O₂ in terms of the rate constants as shown below:

[tex]k_1C_{NO}C_{O_2}=k_2C_{NO_2}C_{O}\\\\C_{O_2}=\frac{k_2C_{NO_2}C_{O}}{k_1C_{NO}}[/tex]

Best regards.

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