Chapter 16, Problem 90A

Interpretation Introduction

Interpretation:

The mass of propene produced in 10.0 min in a volume of 2.50 L needs to be determined, if the rate law is $\text{6}{\text{.22×10}}^{\text{-4}}{\text{s}}^1$ and the concentration of cyclopropane is held at 0.300 mol/L.

Concept introduction:

The rate law for a reaction, $aA\to \text{products}$ id given by the formula,

$rate=k{\left[A\right]}^m$

Where, [A] is the concentration of reactant A, k is the specific rate constant and exponent m is the order of the reactant.

Answer to Problem 90A

The mass of propene produced in 10.0 min in a volume of 2.50 L if the rate law is $\text{6}{\text{.22×10}}^{\text{-4}}{\text{s}}^1$ and the concentration of cyclopropane is held at 0.300 mol/L is $\text{1}\text{.18}\text{g}$.

Explanation of Solution

The conversion of cyclopropane to propene is a first order reaction in cyclopropane. So, value of m = 1. The value of k is given as $\text{6}{\text{.22×10}}^{\text{-4}}{\text{s}}^1$ and the initial concentration of $\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\right]\text{=}\text{0}\text{.0300}\text{mol/L}$.

Using these values, we determine the reaction rate.

$rate=k{\left[A\right]}^m$

$\begin{array}{c}=\left(\text{6}{\text{.22×10}}^{\text{-4}}{\text{s}}^1\right){\left(0.0300\frac{\text{mol}}{\text{L}}\right)}^1\\ =\left(\text{6}{\text{.22×10}}^{\text{-4}}\times 0.0300\right)\frac{\text{mol}}{\text{L}\text{. s}}\\ =1.87\times {10}^{-5}\frac{\text{mol}}{\text{L}\text{. s}}\end{array}$

Therefore, the reaction rate is $1.87\times {10}^{-5}\frac{\text{mol}}{\text{L}\text{. s}}$.

Now, to determine the mass of propene that is produced in 10 minutes in a volume 2.50 L, we will multiply the reaction rate by the time passed and the volume in liters considered.

$\therefore$ propene produced,

$\begin{array}{c}=\left(1.87\times {10}^{-5}\frac{\text{mol}}{\text{L}\text{. s}}\right)\left(10.0\min \times \frac{60s}{1\min }\right)\left(2.50\text{L}\right)\\ =\left(1.87\times {10}^{-5}\times 10.0\times 2.50\right)\text{mol}\\ \text{=}\text{2}\text{.81}\times {\text{10}}^{-2}\text{mol}\end{array}$

Hence, the number of moles of propene that is produced after 10 minutes is $\text{2}\text{.81}\text{×}{\text{10}}^{\text{-2}}\text{mol}$

Now we multiply this value by molar mass of propene, $42.1\frac{\text{g}}{\text{mol}}$

So,

$\begin{array}{c}\text{Mass}\text{of propene produced=(2}\text{.81}\text{×}{\text{10}}^{\text{-2}}\text{mol)}\left(42.1\frac{\text{g}}{\text{mol}}\right)\\ =\left(\text{2}\text{.81}\text{×}{\text{10}}^{\text{-2}}\times 42.1\right)\text{g}\\ \text{=}\text{1}\text{.18}\text{g}\end{array}$

Conclusion

The mass of propene produced is $\text{1}\text{.18}\text{g}$