Medium
From the rate expression for the following reaction,determine its order of reaction and the dimensions of the rate constant.
$(iii)$ $CH_{3}CHO_{(g)} \rightarrow CH_{4(g)} + CO_{(g)} \quad$ Rate $= k[CH_{3}CHO]^{3/2}$
$(iii)$ $CH_{3}CHO_{(g)} \rightarrow CH_{4(g)} + CO_{(g)} \quad$ Rate $= k[CH_{3}CHO]^{3/2}$
(N/A) Given the rate expression: $\text{Rate} = k[CH_{3}CHO]^{3/2}$
$1$. The order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Order of reaction $= \frac{3}{2} = 1.5$
$2$. The dimension of the rate constant $k$ is calculated as:
$k = \frac{\text{Rate}}{[CH_{3}CHO]^{3/2}}$
Substituting the units:
$k = \frac{\text{mol} \ L^{-1} \ s^{-1}}{(\text{mol} \ L^{-1})^{3/2}}$
$k = \frac{\text{mol} \ L^{-1} \ s^{-1}}{\text{mol}^{3/2} \ L^{-3/2}}$
$k = \text{mol}^{(1 - 3/2)} \ L^{(-1 + 3/2)} \ s^{-1}$
$k = \text{mol}^{-1/2} \ L^{1/2} \ s^{-1}$
$1$. The order of reaction is the sum of the powers of the concentration terms in the rate law expression.
Order of reaction $= \frac{3}{2} = 1.5$
$2$. The dimension of the rate constant $k$ is calculated as:
$k = \frac{\text{Rate}}{[CH_{3}CHO]^{3/2}}$
Substituting the units:
$k = \frac{\text{mol} \ L^{-1} \ s^{-1}}{(\text{mol} \ L^{-1})^{3/2}}$
$k = \frac{\text{mol} \ L^{-1} \ s^{-1}}{\text{mol}^{3/2} \ L^{-3/2}}$
$k = \text{mol}^{(1 - 3/2)} \ L^{(-1 + 3/2)} \ s^{-1}$
$k = \text{mol}^{-1/2} \ L^{1/2} \ s^{-1}$
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