Question

For the reaction A + B + 2C → Products; On doubling concentration of A only, rate becomes 2 times. On doubling conc. of B only rate becomes 2.828 times and on doubling concentration of both A and C rate becomes 2 times. Determine overall order.

Answer 1

2.5

Answer 2

Let the rate law for the reaction be given by: Rate $= k [A]^x [B]^y [C]^z$ where $k$ is the rate constant and $x, y, z$ are the orders of the reaction with respect to A, B, and C, respectively. The overall order of the reaction is $x + y + z$.

According to the given information:

1. On doubling the concentration of A only, the rate becomes 2 times. Initial Rate $= k [A]^x [B]^y [C]^z$ New Rate $= k [2A]^x [B]^y [C]^z = k (2^x) [A]^x [B]^y [C]^z = 2^x (\text{Initial Rate})$ Given that New Rate $= 2 \times$ Initial Rate. So, $2^x = 2$, which implies $x = 1$.

2. On doubling the concentration of B only, the rate becomes 2.828 times. Initial Rate $= k [A]^x [B]^y [C]^z$ New Rate $= k [A]^x [2B]^y [C]^z = k (2^y) [A]^x [B]^y [C]^z = 2^y (\text{Initial Rate})$ Given that New Rate $= 2.828 \times$ Initial Rate. So, $2^y = 2.828$. Recognizing that $2.828 \approx 2\sqrt{2} = 2 \times 2^{1/2} = 2^{1+1/2} = 2^{3/2}$. Thus, $2^y = 2^{3/2}$, which implies $y = 3/2 = 1.5$.

3. On doubling the concentration of both A and C, the rate becomes 2 times. We know $x = 1$. Initial Rate $= k [A]^1 [B]^y [C]^z$ New Rate $= k [2A]^1 [B]^y [2C]^z = k (2^1) [A]^1 [B]^y (2^z) [C]^z = 2^{1+z} k [A]^1 [B]^y [C]^z = 2^{1+z} (\text{Initial Rate})$ Given that New Rate $= 2 \times$ Initial Rate. So, $2^{1+z} = 2$. This implies $1 + z = 1$, which gives $z = 0$.

The individual orders are $x=1$, $y=1.5$, and $z=0$. The overall order of the reaction is the sum of the individual orders: Overall Order $= x + y + z = 1 + 1.5 + 0 = 2.5$.