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The rate of the reaction is measured to be 32.0 M / s when [N2] = 0.11 M and [O3] = 0.51 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

Question

The rate of the reaction is measured to be 32.0 M / s when [N2] = 0.11 M and [O3] = 0.51 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

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Solution

To find the rate constant k k , we can use the rate law equation for a reaction. The general form of the rate law is:

Rate=k[N2]m[O3]n \text{Rate} = k [\text{N}_2]^m [\text{O}_3]^n

Given:

  • Rate = 32.0 M/s
  • [N2]=0.11 M[\text{N}_2] = 0.11 \text{ M}
  • [O3]=0.51 M[\text{O}_3] = 0.51 \text{ M}

Assuming the reaction is first order with respect to both N2\text{N}_2 and O3\text{O}_3 (i.e., m=1 m = 1 and n=1 n = 1 ), the rate law simplifies to:

Rate=k[N2][O3] \text{Rate} = k [\text{N}_2] [\text{O}_3]

Now, we can solve for k k :

k=Rate[N2][O3] k = \frac{\text{Rate}}{[\text{N}_2] [\text{O}_3]}

Substitute the given values into the equation:

k=32.0 M/s(0.11 M)(0.51 M) k = \frac{32.0 \text{ M/s}}{(0.11 \text{ M})(0.51 \text{ M})}

Calculate the denominator:

0.11 M×0.51 M=0.0561 M2 0.11 \text{ M} \times 0.51 \text{ M} = 0.0561 \text{ M}^2

Now, divide the rate by this product:

k=32.0 M/s0.0561 M2 k = \frac{32.0 \text{ M/s}}{0.0561 \text{ M}^2}

k≈570.4 s−1 M−1 k \approx 570.4 \text{ s}^{-1} \text{ M}^{-1}

Rounding to 2 significant digits:

k≈570 s−1 M−1 k \approx 570 \text{ s}^{-1} \text{ M}^{-1}

So, the value of the rate constant is approximately 570 s−1 M−1 570 \text{ s}^{-1} \text{ M}^{-1} .

This problem has been solved

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