For the reaction 2NOBr(g) rightarrow 2NO(g) + | vedclass

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Question

(A) Step $1$: Identify the given values: $k = 1.62 	imes 10^{-2} 	ext{ L mol}^{-1} 	ext{ s}^{-1}$ and $[NOBr] = 2 	imes 10^{-3} 	ext{ mol L}^{-1}

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$6.48 	imes 10^{-8} 	ext{ mol L}^{-1} 	ext{ s}^{-1}$

Vedclass · Answer

(A) Step $1$: Identify the given values: $k = 1.62 	imes 10^{-2} 	ext{ L mol}^{-1} 	ext{ s}^{-1}$ and $[NOBr] = 2 	imes 10^{-3} 	ext{ mol L}^{-1}$.Step $2$: Use the rate law expression $r = k[NOBr]^2$.Step $3$: Substitute the values into the equation: $r = (1.62 	imes 10^{-2}) 	imes (2 	imes 10^{-3})^2$.Step $4$: Calculate the square of the concentration: $(2 	imes 10^{-3})^2 = 4 	imes 10^{-6} 	ext{ mol}^2 	ext{ L}^{-2}$.Step $5$: Multiply by the rate constant: $r = 1.62 	imes 10^{-2} 	imes 4 	imes 10^{-6} = 6.48 	imes 10^{-8} 	ext{ mol L}^{-1} 	ext{ s}^{-1}$.

For the reaction $2NOBr(g) \rightarrow 2NO(g) + Br_2(g)$, the rate law is $r = k[NOBr]^2$. If the rate constant $k = 1.62 \times 10^{-2} \text{ L mol}^{-1} \text{ s}^{-1}$ and the concentration of $[NOBr] = 2 \times 10^{-3} \text{ mol L}^{-1}$, what is the rate of reaction?

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