Consider the reaction,,
$2 ICl (g) + H_{2} (g) ⟶ 2 HCl (g) + I_{2} (g)$
At a certain temperature the rate constant is found to be $1.63 \times 10^{- 6} L / {mol - s^{- 1}}_{}$ . What is the overall order of the reaction?

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Third

Second

First

Zero

answer is C.

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Detailed Solution

$\begin{array}{r} \frac{k = 1 .63 \times 10^{- 6} L / mol s^{- 1}}{A ⟶ Product} \\ rate = k [a]^{n} n - order \end{array}$

$k = \frac{rate}{[A]^{n}} = \frac{M / \sec}{M^{n}}$

$\begin{array}{l} k = M / \sec \\ = Mol / L - \sec . \end{array}$

$1 st k = L / \sec$

$\begin{array}{r} 2^{nd} n = 2 k = \frac{M / \sec}{M^{2}} = M^{- 1} \sec^{- 1} \\ = L / mol \sec^{- 1} \end{array}$

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