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Step-by-Step Solution
Step 1: Write down the form of the rate law
Since the given reaction is elementary:
2A + B_2 \to 2AB,
the rate law can be written directly from its stoichiometry, giving:
\text{Rate} = k [A]^2 [B_2].
Step 2: Express initial concentrations in terms of volume
Let the initial concentrations of A and B2 be
[A]_i = \frac{a}{V} \quad \text{and} \quad [B_2]_i = \frac{b}{V},
where a and b are the amounts of A and B2 respectively, and V is the volume of the vessel.
Step 3: Calculate the initial rate
The initial rate, r_1 , is:
r_1 = k \left(\frac{a}{V}\right)^2 \left(\frac{b}{V}\right).
Step 4: Determine the effect of reducing the volume by a factor of 3
If the volume is reduced by a factor of 3, the new volume is
\frac{V}{3}.
Hence, the new concentrations become:
[A]_{\text{new}} = \frac{a}{\frac{V}{3}} = \frac{3a}{V}, \quad [B_2]_{\text{new}} = \frac{b}{\frac{V}{3}} = \frac{3b}{V}.
Step 5: Calculate the new rate
The new rate, r_2 , is:
r_2 = k \left(\frac{3a}{V}\right)^2 \left(\frac{3b}{V}\right).
Step 6: Find the factor by which the rate increases
Divide r_2 by r_1 to see the change in rate:
\frac{r_2}{r_1}
= \frac{k \left(\frac{3a}{V}\right)^2 \left(\frac{3b}{V}\right)}{k \left(\frac{a}{V}\right)^2 \left(\frac{b}{V}\right)}
= 3^2 \times 3 = 3^3 = 27.
Thus, the rate increases by a factor of 27.
Step 7: State the final answer
The rate of reaction increases by a factor of 27.
