For the reaction $2\text{NOBr}(g) \to 2\text{NO}_2(g) + \text{Br}_2(g)$, rate = $k[\text{NOBr}]^2$. If rate = $6.5 \times 10^{-6}$ mol L⁻¹ s⁻¹ when [NOBr] = $2 \times 10^{-3}$ M, then $k$ equals:
A$6.5 \times 10^{-6}$ s⁻¹
B$1.3 \times 10^{-9}$ M s⁻¹
C$3.25 \times 10^{-3}$ M⁻¹ s⁻¹
D$1.625$ M⁻¹ s⁻¹
Answer & Solution
Correct answer: D. $1.625$ M⁻¹ s⁻¹
$k = \text{rate}/[\text{NOBr}]^2 = (6.5\times10^{-6})/(2\times10^{-3})^2 = 6.5\times10^{-6}/4\times10^{-6} = 1.625$ M⁻¹ s⁻¹.
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