Question

1.00 points 3 attempts left Check my work 2 Enter your answer in the provided box. The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.800-s) at 10°C. Starting with a concentration of 0.86 M, calculate the concentration of NOBr Report problem Hint Solution Guided Solution after 67 s. Be sure to report your answer to the correct number of significant figures. References eBook &Resources

Answer 1

The rate law for this second-order reaction is expressed as -

$-\frac{1}{2} \frac{d[A]}{dt}=k[A]^2 \\ \\ \frac{d[A]}{dt}=-2k[A]^2$

This equation, after integration, can be rearranged as -

$\frac{1}{[A]} = \frac{1}{[A]_0} + 2kt$

where, A is concentration of the reactant (here, NOBr) at any time t, A₀ is the initinal concentration of reactant and k is the rate constant.

Using the data available, we can calculate conc. of NOBr after 67s as follows -

$\frac{1}{[A]} = \frac{1}{0.86 M} + 2 \times 0.80\, M^{-1}s^{-1} \times 67 s \\ \\ \frac{1}{[A]} = 108.36 \\ \\ {[A]} = 0.0092 M \approx 0.01 M$

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