Medium
The three experimental data sets for determining the differential rate of the reaction $2 NO_{(g)} + Cl_{2_{(g)}} \rightarrow 2 NOCl_{(g)}$ at a definite temperature are given below. (Note: The data table was missing in the input,assuming standard values for this reaction: $Exp 1: [NO]=0.1, [Cl_2]=0.1, Rate=0.18$; $Exp 2: [NO]=0.1, [Cl_2]=0.2, Rate=0.36$; $Exp 3: [NO]=0.2, [Cl_2]=0.1, Rate=0.72$).
$(a)$ Derive the differential rate law of the reaction.
$(b)$ Calculate the order of the reaction.
$(c)$ Calculate the value of the rate constant.
$(a)$ Derive the differential rate law of the reaction.
$(b)$ Calculate the order of the reaction.
$(c)$ Calculate the value of the rate constant.
(C) The differential rate law is given by: $-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[Cl_2]}{dt} = k[NO]^x[Cl_2]^y$.
$(b)$ Comparing $Exp 1$ and $Exp 2$: Keeping $[NO]$ constant,doubling $[Cl_2]$ doubles the rate,so $y=1$. Comparing $Exp 1$ and $Exp 3$: Keeping $[Cl_2]$ constant,doubling $[NO]$ quadruples the rate,so $x=2$. The rate law is $Rate = k[NO]^2[Cl_2]^1$. The order of reaction $= 2 + 1 = 3$.
$(c)$ Using $Exp 1$: $0.18 = k(0.1)^2(0.1)^1$ $\Rightarrow 0.18 = k(0.001)$ $\Rightarrow k = 180 \ L^2 \ mol^{-2} \ s^{-1}$.
$(b)$ Comparing $Exp 1$ and $Exp 2$: Keeping $[NO]$ constant,doubling $[Cl_2]$ doubles the rate,so $y=1$. Comparing $Exp 1$ and $Exp 3$: Keeping $[Cl_2]$ constant,doubling $[NO]$ quadruples the rate,so $x=2$. The rate law is $Rate = k[NO]^2[Cl_2]^1$. The order of reaction $= 2 + 1 = 3$.
$(c)$ Using $Exp 1$: $0.18 = k(0.1)^2(0.1)^1$ $\Rightarrow 0.18 = k(0.001)$ $\Rightarrow k = 180 \ L^2 \ mol^{-2} \ s^{-1}$.
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