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)
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.
