Question

Consider the following single step reaction in gas phase at constant temperature.
\(2A_{(g)} + B_{(g)} → C_{(g)}\)
The initial rate of the reaction is recorded as $r_1$ when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate $r_2$ is recorded when the pressure of C becomes 0.5 atm. The ratio $r_1 : r_2$ is _______ $\times 10^1$. (Nearest integer)

Answer 1

Correct Answer: 315

The rate law for the reaction is:
\[r = k[A]^2[B],\]
where \([A]\), \([B]\), and \([C]\) represent the partial pressures of the reactants and product.
Step 1: Initial conditions at \(r_1\):
\([A] = 1.5 \, \text{atm}, \, [B] = 0.7 \, \text{atm}\).  
\[r_1 = k[1.5]^2[0.7].\]
Step 2: Conditions when \(r_2\) is measured  
When the pressure of \([C]\) is \(0.5 \, \text{atm}\), due to stoichiometry:
\[2A_{(g)} + B_{(g)} \rightarrow C_{(g)},\]
the change in \([C]\) corresponds to:
\[\Delta[C] = 0.5 \, \text{atm}.\]
This means:
\[\Delta[A] = 2 \times \Delta[C] = 2 \times 0.5 = 1.0 \, \text{atm}.\]
\[\Delta[B] = \Delta[C] = 0.5 \, \text{atm}.\]
The remaining pressures of \(A\) and \(B\) are:
\[[A] = 1.5 - 1.0 = 0.5 \, \text{atm},\]
\[[B] = 0.7 - 0.5 = 0.2 \, \text{atm}.\]
At \(r_2\):
\[r_2 = k[0.5]^2[0.2].\]
Step 3: Calculate the ratio \(\frac{r_1}{r_2}\):  
The ratio of the rates is:
\[\frac{r_1}{r_2} = \frac{k[1.5]^2[0.7]}{k[0.5]^2[0.2]}.\]
Simplify:
\[\frac{r_1}{r_2} = \frac{(1.5)^2(0.7)}{(0.5)^2(0.2)} = \frac{2.25 \times 0.7}{0.25 \times 0.2}.\]
\[\frac{r_1}{r_2} = \frac{1.575}{0.05} = 31.5 \times 10^{-1}.\]
Step 4: Nearest integer:
\[x = 315.\]
Final Answer: 315.