Question

the rate of a certain reaction is given by the following rate law: rate = kn2^3o3. use this information to answer the questions below. what is the reaction order in n2? what is the reaction order in o3? what is the overall reaction order? at a certain concentration of n2 and o3, the initial rate of the reaction is 56.0 m/s. what would the initial rate of the reaction be if the concentration of n2 were doubled? be sure your answer has the correct number of significant digits. the rate of the reaction is measured to be 2.0×10^3 m/s when n2 = 1.7 m and o3 = 1.5 m. calculate the value of the rate constant. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Determine reaction orders

The rate - law is given by rate = k$[N_2]^3[O_3]$. The exponent of $N_2$ in the rate - law is the reaction order with respect to $N_2$, and the exponent of $O_3$ is the reaction order with respect to $O_3$.
The reaction order in $N_2$ is 3 and the reaction order in $O_3$ is 1.

Step2: Calculate overall reaction order

The overall reaction order is the sum of the reaction orders with respect to each reactant. So, overall reaction order=3 + 1=4.

Step3: Calculate the rate constant k

We know that rate = k$[N_2]^3[O_3]$. Given rate = 2.0×$10^3$ M/s, $[N_2]$ = 1.7 M, and $[O_3]$ = 1.5 M.
We can re - arrange the rate - law to solve for k: k=$\frac{rate}{[N_2]^3[O_3]}$.
Substitute the given values: k=$\frac{2.0\times10^3}{(1.7)^3\times1.5}$.
First, calculate $(1.7)^3=1.7\times1.7\times1.7 = 4.913$.
Then, $(1.7)^3\times1.5=4.913\times1.5 = 7.3695$.
k=$\frac{2.0\times10^3}{7.3695}\approx271.4$ $M^{-3}s^{-1}$.

Step4: Determine the effect of doubling $N_2$ concentration

If the concentration of $N_2$ is doubled while $[O_3]$ is constant, the new rate $rate_{new}$ and the original rate $rate_{old}$ are related as follows:
$rate_{old}$ = k$[N_2]_{old}^3[O_3]_{old}$ and $rate_{new}$ = k$[N_2]_{new}^3[O_3]_{new}$. Since $[O_3]_{new}=[O_3]_{old}$ and $[N_2]_{new} = 2[N_2]_{old}$, then $rate_{new}$ = k$(2[N_2]_{old})^3[O_3]_{old}=8k[N_2]_{old}^3[O_3]_{old}=8rate_{old}$. So the rate would be multiplied by 8.

Answer:

k = 271.4 $M^{-3}s^{-1}$
If $N_2$ concentration is doubled, rate changes by a factor of 8
Overall reaction order: 4
Reaction order in $O_3$: 1
Reaction order in $N_2$: 3