Consider the elementary reaction A_{(g)} + B_ | vedclass

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(B) For an elementary reaction $A_{(g)} + B_{(g)} \rightarrow C_{(g)} + D_{(g)}$,the rate law is given by $R_1 = k[A][B]$.Since concentration is moles

Vedclass · Accepted Answer

$9$

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(B) For an elementary reaction $A_{(g)} + B_{(g)} \rightarrow C_{(g)} + D_{(g)}$,the rate law is given by $R_1 = k[A][B]$.Since concentration is moles per unit volume,$R_1 = k \left( \frac{n_A}{V} \right) \left( \frac{n_B}{V} \right) = k \frac{n_A n_B}{V^2}$.When the volume is reduced to $\frac{V}{3}$,the new concentration becomes $3$ times the initial concentration.$R_2 = k [3A][3B] = 9k[A][B]$.Therefore,$R_2 = 9R_1$,which means the rate becomes $9$ times the original rate.Thus,$x = 9$.

Consider the elementary reaction $A_{(g)} + B_{(g)} \rightarrow C_{(g)} + D_{(g)}$. If the volume of the reaction mixture is suddenly reduced to $\frac{1}{3}$ of its initial volume,the reaction rate will become '$x$' times the original reaction rate. The value of $x$ is:

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