Question

Imagine that you run this experiment and that the initial change in pressure is 0.0023 atm/s. what is the initial reaction rate? the temperature of the water bath is 20.0 oc.

Answer 1

$\text{reaction rate} = 9.6 \times 10^{-5} \; \text{mol}\cdot\text{L}^{-1} \cdot \text{s}^{-1}$

Explanation:

Concentration $c = n / V$;

The reaction see a $0.0023 \; \text{atm}$ change in pressure per second. By the ideal gas law, this quantity would correspond to a change in concentration, $n/V$ of

$c = n / V = P / (n \cdot R) = 0.0023 / ((273.15 + 20) \times 0.0821 ) = 9.6 \times 10^{-5} \; \text{mol}$

where the ideal gas constant $R = 0.0821 \; \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.

By definition, $\text{reaction rate} = \Delta \text{concentration} / \Delta\text{time}$

Therefore, for this particular reaction

$\text{rate} = 9.6 \times 10^{-5} \; \text{mol} / (1 \; \text{s} ) = 9.6 \times 10^{-5} \; \text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1}$