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Chemistry Question on Chemical Kinetics

Consider the following single step reaction in gas phase at constant temperature.
2A(g)+B(g)→C(g)2A_{(g)} + B_{(g)} → C_{(g)}
The initial rate of the reaction is recorded as r1r_1 when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate r2r_2 is recorded when the pressure of C becomes 0.5 atm. The ratio r1:r2r_1 : r_2 is _______ ×101\times 10^1. (Nearest integer)

Answer

The rate law for the reaction is:
r=k[A]2[B],r = k[A]^2[B],
where [A][A], [B][B], and [C][C] represent the partial pressures of the reactants and product.
Step 1: Initial conditions at r1r_1:
[A]=1.5 atm, [B]=0.7 atm[A] = 1.5 \, \text{atm}, \, [B] = 0.7 \, \text{atm}.
r1=k[1.5]2[0.7].r_1 = k[1.5]^2[0.7].
Step 2: Conditions when r2r_2 is measured
When the pressure of [C][C] is 0.5 atm0.5 \, \text{atm}, due to stoichiometry:
2A(g)+B(g)→C(g),2A_{(g)} + B_{(g)} \rightarrow C_{(g)},
the change in [C][C] corresponds to:
Δ[C]=0.5 atm.\Delta[C] = 0.5 \, \text{atm}.
This means:
Δ[A]=2×Δ[C]=2×0.5=1.0 atm.\Delta[A] = 2 \times \Delta[C] = 2 \times 0.5 = 1.0 \, \text{atm}.
Δ[B]=Δ[C]=0.5 atm.\Delta[B] = \Delta[C] = 0.5 \, \text{atm}.
The remaining pressures of AA and BB are:
[A]=1.5−1.0=0.5 atm,[A] = 1.5 - 1.0 = 0.5 \, \text{atm},
[B]=0.7−0.5=0.2 atm.[B] = 0.7 - 0.5 = 0.2 \, \text{atm}.
At r2r_2:
r2=k[0.5]2[0.2].r_2 = k[0.5]^2[0.2].
Step 3: Calculate the ratio r1r2\frac{r_1}{r_2}:
The ratio of the rates is:
r1r2=k[1.5]2[0.7]k[0.5]2[0.2].\frac{r_1}{r_2} = \frac{k[1.5]^2[0.7]}{k[0.5]^2[0.2]}.
Simplify:
r1r2=(1.5)2(0.7)(0.5)2(0.2)=2.25×0.70.25×0.2.\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}.
r1r2=1.5750.05=31.5×10−1.\frac{r_1}{r_2} = \frac{1.575}{0.05} = 31.5 \times 10^{-1}.
Step 4: Nearest integer:
x=315.x = 315.
Final Answer: 315.