Question

For a gaseous elementary reaction $2A + B \rightarrow C$

If the volume of the reaction vessel is reduced to one third of the initial volume, the reaction rate with respect to original rate will the

Answer 1

27

Answer 2

The given reaction is an elementary gaseous reaction: $2A + B \rightarrow C$

For an elementary reaction, the rate law is determined directly from the stoichiometric coefficients of the reactants. So, the rate law for this reaction is: Rate = $k[A]^2[B]$

Let the initial volume of the reaction vessel be $V_1$. Let the initial concentrations of A and B be $[A]_1$ and $[B]_1$, respectively. The initial rate of the reaction, $R_1$, is: $R_1 = k[A]_1^2[B]_1$

The volume of the reaction vessel is reduced to one third of the initial volume. So, the new volume, $V_2 = \frac{1}{3}V_1$.

For a given amount of substance, concentration is inversely proportional to volume ($C = \frac{n}{V}$). If the volume is reduced to one third, the concentration of each gaseous reactant will become three times its original concentration. New concentration of A: $[A]_2 = 3[A]_1$ New concentration of B: $[B]_2 = 3[B]_1$

Now, let's calculate the new rate of the reaction, $R_2$, using these new concentrations: $R_2 = k[A]_2^2[B]_2$ Substitute the new concentrations: $R_2 = k(3[A]_1)^2(3[B]_1)$ $R_2 = k(9[A]_1^2)(3[B]_1)$ $R_2 = 27 k[A]_1^2[B]_1$

To find the reaction rate with respect to the original rate, we compare $R_2$ with $R_1$: Since $R_1 = k[A]_1^2[B]_1$, we can substitute $R_1$ into the expression for $R_2$: $R_2 = 27 R_1$

Therefore, the reaction rate will be 27 times the original rate.

Explanation of the solution:

1. Determine the rate law: For an elementary reaction $2A + B \rightarrow C$, the rate law is Rate = $k[A]^2[B]$.
2. Initial Rate: Let initial concentrations be $[A]$ and $[B]$. Initial rate $R_1 = k[A]^2[B]$.
3. Effect of Volume Change on Concentration: Reducing the volume to one-third means concentrations become three times their original values (since concentration is inversely proportional to volume). So, new concentrations are $3[A]$ and $3[B]$.
4. New Rate: Substitute new concentrations into the rate law: $R_2 = k(3[A])^2(3[B]) = k(9[A]^2)(3[B]) = 27k[A]^2[B]$.
5. Ratio of Rates: Compare $R_2$ to $R_1$: $R_2 = 27 R_1$. The new rate is 27 times the original rate.