The elementary reaction 2SO_{2(g)} + O_{2(g)} | vedclass

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(D) For an elementary reaction,the rate law is determined by the stoichiometry of the reaction. Given reaction: $2SO_{2(g)} + O_{2(g)} 	o 2SO_{3(g)}$

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$8 : 1$

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(D) For an elementary reaction,the rate law is determined by the stoichiometry of the reaction. Given reaction: $2SO_{2(g)} + O_{2(g)} 	o 2SO_{3(g)}$ Rate law: $r = k[SO_2]^2[O_2]$ Let the number of moles of $SO_2$ be $n_1$ and $O_2$ be $n_2$. Concentration is given by $C = n/V$. $r = k(n_1/V)^2(n_2/V) = k(n_1^2 n_2) / V^3$ Since $r \propto 1/V^3$,the ratio of rates in two vessels of volumes $V_1 = 1 \ dm^3$ and $V_2 = 2 \ dm^3$ is: $r_1 / r_2 = (V_2 / V_1)^3 = (2 / 1)^3 = 8 / 1$ Therefore,the ratio is $8 : 1$.

The elementary reaction $2SO_{2(g)} + O_{2(g)} \to 2SO_{3(g)}$ is carried out in a $1 \ dm^3$ vessel and a $2 \ dm^3$ vessel separately. The ratio of the reaction velocities will be

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