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 * 10^(-5) \; \text{mol}\cdot\text{L}^(-1) \cdot \text{s}^(-1)`

Step-by-step 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) * 0.0821 ) = 9.6 * 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 * 10^(-5) \; \text{mol} / (1 \; \text{s} ) = 9.6 * 10^(-5) \; \text{mol} \cdot \text{L}^(-1) \cdot \text{s}^(-1)`