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Question: A follows parallel path, first order reaction giving B and C as If initial concentration of A is 0...

A follows parallel path, first order reaction giving B and C as

If initial concentration of A is 0.25 M, calculate the concentration of C after 5 hours of reaction. [Given : k₁ = 1.5 × 10−510^{-5} s−1s^{-1}, k₂ = 5 × 10−610^{-6} s−1s^{-1}]

A

0.0209 M

B

0.0150 M

C

0.0050 M

D

0.0300 M

Answer

0.0209 M

Explanation

Solution

The problem describes a first-order reaction where reactant A decomposes via two parallel paths to form products B and C. The reactions are given by:

  1. 5A→k1B5\text{A} \xrightarrow{k_1} \text{B}
  2. 5A→k22C5\text{A} \xrightarrow{k_2} 2\text{C}

The rate of disappearance of A is the sum of the rates of consumption of A in each pathway. Assuming each pathway is first-order with respect to A, the rate of reaction for the first pathway is r1=k1[A]r_1 = k_1[\text{A}] and for the second pathway is r2=k2[A]r_2 = k_2[\text{A}]. The rate of disappearance of A is given by: −d[A]dt=5×r1+5×r2=5k1[A]+5k2[A]=5(k1+k2)[A]-\frac{d[\text{A}]}{dt} = 5 \times r_1 + 5 \times r_2 = 5 k_1 [\text{A}] + 5 k_2 [\text{A}] = 5(k_1 + k_2)[\text{A}] The observed rate constant for the disappearance of A is kobs=5(k1+k2)k_{obs} = 5(k_1 + k_2).

The concentration of A at time tt is given by: [A]t=[A]0e−kobst[\text{A}]_t = [\text{A}]_0 e^{-k_{obs} t}

For the formation of C, the second reaction is 5A→k22C5\text{A} \xrightarrow{k_2} 2\text{C}. The rate of this reaction is r2=k2[A]r_2 = k_2[\text{A}]. This rate defines the speed of the reaction. The rate of formation of 2C is equal to the rate of reaction r2r_2, so d[2C]dt=k2[A]\frac{d[2\text{C}]}{dt} = k_2[\text{A}]. Since the stoichiometry shows that 2 moles of C are formed for every 5 moles of A consumed in this pathway, the rate of formation of C is twice the rate of formation of 2C: d[C]dt=2×d[2C]dt=2k2[A]\frac{d[\text{C}]}{dt} = 2 \times \frac{d[2\text{C}]}{dt} = 2 k_2 [\text{A}] To find the concentration of C at time tt, we integrate this rate equation: [C]t=∫0td[C]dtdt=∫0t2k2[A]tdt[\text{C}]_t = \int_0^t \frac{d[\text{C}]}{dt} dt = \int_0^t 2 k_2 [\text{A}]_t dt Substitute [A]t=[A]0e−kobst[\text{A}]_t = [\text{A}]_0 e^{-k_{obs} t}: [C]t=∫0t2k2[A]0e−kobstdt=2k2[A]0∫0te−kobstdt[\text{C}]_t = \int_0^t 2 k_2 [\text{A}]_0 e^{-k_{obs} t} dt = 2 k_2 [\text{A}]_0 \int_0^t e^{-k_{obs} t} dt [C]t=2k2[A]0[e−kobst−kobs]0t=2k2[A]0(1−e−kobstkobs)[\text{C}]_t = 2 k_2 [\text{A}]_0 \left[ \frac{e^{-k_{obs} t}}{-k_{obs}} \right]_0^t = 2 k_2 [\text{A}]_0 \left( \frac{1 - e^{-k_{obs} t}}{k_{obs}} \right)

Given values: [A]0=0.25[\text{A}]_0 = 0.25 M k1=1.5×10−5 s−1k_1 = 1.5 \times 10^{-5} \text{ s}^{-1} k2=5×10−6 s−1k_2 = 5 \times 10^{-6} \text{ s}^{-1} t=5t = 5 hours =5×3600 s=18000 s= 5 \times 3600 \text{ s} = 18000 \text{ s}

Calculate kobsk_{obs}: kobs=5(k1+k2)=5(1.5×10−5 s−1+5×10−6 s−1)k_{obs} = 5(k_1 + k_2) = 5(1.5 \times 10^{-5} \text{ s}^{-1} + 5 \times 10^{-6} \text{ s}^{-1}) kobs=5(1.5×10−5 s−1+0.5×10−5 s−1)=5(2.0×10−5 s−1)=1.0×10−4 s−1k_{obs} = 5(1.5 \times 10^{-5} \text{ s}^{-1} + 0.5 \times 10^{-5} \text{ s}^{-1}) = 5(2.0 \times 10^{-5} \text{ s}^{-1}) = 1.0 \times 10^{-4} \text{ s}^{-1}

Calculate kobstk_{obs} t: kobst=(1.0×10−4 s−1)×(18000 s)=1.8k_{obs} t = (1.0 \times 10^{-4} \text{ s}^{-1}) \times (18000 \text{ s}) = 1.8

Calculate e−kobste^{-k_{obs} t}: e−1.8≈0.1653e^{-1.8} \approx 0.1653

Calculate [C]t[\text{C}]_t: [C]t=2×(5×10−6 s−1)×(0.25 M)×(1−0.16531.0×10−4 s−1)[\text{C}]_t = 2 \times (5 \times 10^{-6} \text{ s}^{-1}) \times (0.25 \text{ M}) \times \left( \frac{1 - 0.1653}{1.0 \times 10^{-4} \text{ s}^{-1}} \right) [C]t=(10×10−6)×0.25×(0.83471.0×10−4)[\text{C}]_t = (10 \times 10^{-6}) \times 0.25 \times \left( \frac{0.8347}{1.0 \times 10^{-4}} \right) [C]t=2.5×10−6×0.8347×104[\text{C}]_t = 2.5 \times 10^{-6} \times 0.8347 \times 10^4 [C]t=2.08675×10−2 M[\text{C}]_t = 2.08675 \times 10^{-2} \text{ M} [C]t≈0.0209 M[\text{C}]_t \approx 0.0209 \text{ M}