Q17P - Chemistry: Molecular Nature Of Matter And Change

Question

Express the rate of reaction in terms of the change in concentration of each of the reactants and products: 

$\mathbf{2}\mathbf{D}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{E}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{F}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{G}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{H}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

When [D] is decreasing at 0.1 mol/L.s, how fast is [H] increasing?

Step-by-Step Solution

Answer

The rate of reaction  can be represented in terms of reactant disappearance or product appearance as follows:

$$\begin{gathered} \mathrm{reaction} \mathrm{rate}=\frac{1}{2}\frac{∆\left[\mathrm{D}\right]}{∆\mathrm{t}} \\ =-\frac{1}{3}\frac{∆\left[\mathrm{E}\right]}{∆\mathrm{t}} \\ =\frac{∆\left[\mathrm{F}\right]}{∆\mathrm{t}} \\ =\frac{1}{2}\frac{∆\left[\mathrm{G}\right]}{∆\mathrm{t}} \\ =\frac{∆\left[\mathrm{H}\right]}{∆\mathrm{t}} \end{gathered}$$

When [D] is decreasing at 0.1 mol/L.s, Thus, H is increasing at a rate of 0.05 mol/L.s.

Step 1: The rate of reaction in terms of the change in concentration

The rate of reaction can be represented in terms of reactant disappearance or product appearance as follows:

$$\begin{gathered} \mathrm{reaction} \mathrm{rate}=\frac{1}{2}\frac{∆\left[\mathrm{D}\right]}{∆\mathrm{t}} \\ =-\frac{1}{3}\frac{∆\left[\mathrm{E}\right]}{∆\mathrm{t}} \\ =\frac{∆\left[\mathrm{F}\right]}{∆\mathrm{t}} \\ =\frac{1}{2}\frac{∆\left[\mathrm{G}\right]}{∆\mathrm{t}} \\ =\frac{∆\left[\mathrm{H}\right]}{∆\mathrm{t}} \end{gathered}$$

Where, the inverse of the fraction gives the stoichiometric coefficients of the molecules.

Step 2: Determine the rate of appearance of H

One molecule of H arises for every two molecules of D that vanish. The pace at which H appears can be represented as follows:

$\mathrm{Rate} \mathrm{of} \mathrm{appearance} \mathrm{of} \mathrm{H}=\frac{∆\left[\mathrm{H}\right]}{∆\mathrm{t}}$

Determine the rate of appearance of H:

$$\begin{gathered} -\frac{1}{2}\frac{∆\left[\mathrm{D}\right]}{∆\mathrm{t}}=\frac{∆\left[\mathrm{H}\right]}{∆\mathrm{t}} \\ \frac{∆\left[\mathrm{H}\right]}{∆\mathrm{t}}=-\frac{1}{2}\times \left(0.1\mathrm{mol}/\mathrm{Ls}\right) \\ =0.05\mathrm{mol}/\mathrm{Ls} \end{gathered}$$

Thus, H is increasing at a rate of 0.05 mol/L.s.