How can you determine the rate law of the following reaction?
$2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$
$2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$
(N/A) The rate law is determined experimentally by measuring the initial rate of the reaction at different initial concentrations of the reactants.
$1$. Keep the concentration of one reactant constant and vary the concentration of the other reactant to observe the effect on the reaction rate.
$2$. For the reaction $2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$,if the concentration of $NO$ is doubled while keeping $[O_2]$ constant,the rate increases by a factor of $4$,indicating the order with respect to $NO$ is $2$.
$3$. If the concentration of $O_2$ is doubled while keeping $[NO]$ constant,the rate doubles,indicating the order with respect to $O_2$ is $1$.
$4$. Thus,the rate law is expressed as: $\text{Rate} = k[NO]^2[O_2]^1$.
$1$. Keep the concentration of one reactant constant and vary the concentration of the other reactant to observe the effect on the reaction rate.
$2$. For the reaction $2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$,if the concentration of $NO$ is doubled while keeping $[O_2]$ constant,the rate increases by a factor of $4$,indicating the order with respect to $NO$ is $2$.
$3$. If the concentration of $O_2$ is doubled while keeping $[NO]$ constant,the rate doubles,indicating the order with respect to $O_2$ is $1$.
$4$. Thus,the rate law is expressed as: $\text{Rate} = k[NO]^2[O_2]^1$.
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| $0.05$ | $0.05$ | $1.2 \times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4 \times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2 \times 10^{-3}$ |
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