Question

The rate constant for the second-order reaction !s 0. 54 m-1 s-1 at 300°c. how long (in seconds) would 1t take for the concentration of n02 to decrease from 0. 62 m to 0. 28 m?

Answer 1

The answer is 3.63. seconds.

Second order reaction is the reaction in which the rate of reaction depends on either the concentration of two reactant species or on the two times the concentration of single reactant species.

What is the integrated rate law for the second-order reaction?

• The integrated rate law that relates the concentration, time and rate constant for the second-order reaction is:

$\frac{1}{[A]} =\frac{1}{[A]_{0} } +kt$

Where

$\[\begin{array}{l}{\rm{[A] - concentration\ of\ reactant\ A\ at\ time\ t}}\\{{\rm{[A]}}_0}{\rm{ - initial\ concentration\ of\ reactant\ A}}\\{\rm{t - time}}\\{\rm{k - rate\ constant}}\end{array}\]$

• Now, in the given question,

k = $0.54 m^{-1}. s^{-1}$

$[NO_{2} ]= 0.62\ M$

$[NO_{2} ]_{0} = 0.28\ M$

• Thus, using the rate law, the time is calculated as-

$\frac{1}{0.28\ M} =\frac{1}{0.62\ M } +(0.54 m^{-1}.s^{-1}) t\\\\(0.54 m^{-1}.s^{-1}) t= \frac{1}{0.28\ M} -\frac{1}{0.62\ M } = 1.959$

Therefore,

$t =\frac{1.959}{0.54} = 3.63\ seconds$

• Hence, the it would take 3.63 seconds for the concentration of $NO_{2}$ to decrease from 0.62 M to 0.28 M if the reaction is second order.

To learn more about second order reaction visit:

brainly.com/question/17217385

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