Question

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] =

Answer 1

`C_(O_2)=(k_2C_(NO_2)C_(O))/(k_1C_(NO))`

Step-by-step explanation:

Hello,

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

`r=-k_1C_(NO)C_(O_2)+k_2C_(NO_2)C_(O)`

Nevertheless, at equilibrium, the rate becomes zero:

`0=-k_1C_(NO)C_(O_2)+k_2C_(NO_2)C_(O)`

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

`k_1C_(NO)C_(O_2)=k_2C_(NO_2)C_(O)\\\\C_(O_2)=(k_2C_(NO_2)C_(O))/(k_1C_(NO))`

Best regards.