For the gaseous reaction represented by the equation: 2\text{A}(g) + \text{B}(g) \rightarr… — JAMB UTME

For the gaseous reaction represented by the equation:

2\text{A}(g) + \text{B}(g) \rightarrow \text{C}(g) + 3\text{D}(g)

the quantity of substance

\text{A}

present in a

2.0\text{ dm}^3

reaction vessel decreases from

0.80\text{ mol}

0.32\text{ mol}

40\text{ s}

. What is the average rate of formation of product

\text{D}

\text{mol dm}^{-3}\text{ s}^{-1}

Answer: 0.009 mol dm^-3 s^-1

The average rate of formation of product D is

0.009\text{ mol dm}^{-3}\text{ s}^{-1}

The change in concentration of reactant A over

40\text{ s}

\frac{0.48\text{ mol}}{2.0\text{ dm}^3} = 0.24\text{ mol dm}^{-3}

. The rate of consumption of A is

\frac{0.24}{40} = 0.006\text{ mol dm}^{-3}\text{ s}^{-1}

. Because

2\text{ moles}

of A produce

3\text{ moles}

of D, the rate of formation of D is

\frac{3}{2} \times 0.006 = 0.009\text{ mol dm}^{-3}\text{ s}^{-1}

Calculate the change in concentration of reactant A during the time interval.

\Delta [A] = \frac{0.80\text{ mol} - 0.32\text{ mol}}{2.0\text{ dm}^3} = 0.24\text{ mol dm}^{-3}

Concentration is moles per unit volume.

Calculate the rate of consumption of reactant A per unit time.

\text{Rate}_A = -\frac{\Delta [A]}{\Delta t} = \frac{0.24\text{ mol dm}^{-3}}{40\text{ s}} = 0.006\text{ mol dm}^{-3}\text{ s}^{-1}

Reaction rate is defined as the change in concentration over elapsed time.

Use the stoichiometric coefficients from the balanced equation to calculate the rate of formation of D.

\text{Rate}_D = \frac{3}{2} \times \text{Rate}_A = \frac{3}{2} \times 0.006 = 0.009\text{ mol dm}^{-3}\text{ s}^{-1}

According to the balanced chemical equation,

2

moles of A are consumed for every

3

moles of D formed, so

\frac{1}{3}\text{Rate}_D = \frac{1}{2}\text{Rate}_A

Stoichiometric relationship between rates of consumption of reactants and rates of formation of products

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