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Question: A substance 'A' undergoes reactions by following three different parallel path : $A \xrightarrow{K_1...

A substance 'A' undergoes reactions by following three different parallel path : A→K1B;K1=5×10−2sec−1A \xrightarrow{K_1} B; K_1 = 5 \times 10^{-2} sec^{-1}

2A→K2C;K2=3×10−2sec−12A \xrightarrow{K_2} C; K_2 = 3 \times 10^{-2} sec^{-1}

3A→K3D;K3=5×10−3sec−13A \xrightarrow{K_3} D; K_3 = 5 \times 10^{-3} sec^{-1}

Calculate average life time of substance (in sec.) [K1,K2,K3K_1, K_2, K_3 are rate constants for respective reactions]

A

100085\frac{1000}{85}

B

10085\frac{100}{85}

C

8

D

4

Answer

100085\frac{1000}{85}

Explanation

Solution

The substance 'A' undergoes three parallel reactions:

  1. A→K1BA \xrightarrow{K_1} B with rate constant K1=5×10−2s−1K_1 = 5 \times 10^{-2} s^{-1}
  2. 2A→K2C2A \xrightarrow{K_2} C with rate constant K2=3×10−2s−1K_2 = 3 \times 10^{-2} s^{-1}
  3. 3A→K3D3A \xrightarrow{K_3} D with rate constant K3=5×10−3s−1K_3 = 5 \times 10^{-3} s^{-1}

The units of the rate constants are given as s−1s^{-1} for all three reactions. For a first-order reaction, the rate constant has units of s−1s^{-1}. For a second-order reaction, the rate constant has units of M−1s−1M^{-1}s^{-1} (or concentration−1s−1^{-1}s^{-1}). For a third-order reaction, the rate constant has units of M−2s−1M^{-2}s^{-1} (or concentration−2s−1^{-2}s^{-1}).

The fact that all given rate constants have the unit s−1s^{-1}, and the question asks for a single value of average lifetime (which for reactions other than first order depends on initial concentration), strongly suggests that all three parallel reactions are treated as first-order processes with respect to A, with the given rate constants. The stoichiometry given (2A→C2A \rightarrow C and 3A→D3A \rightarrow D) likely represents the overall stoichiometry but not the rate law order, or the reactions are pseudo-first-order under conditions not explicitly stated (e.g., excess of other reactants, which are not mentioned). Assuming the rate laws are first order in A with the given constants is the most consistent interpretation given the units and the nature of the options.

So, the rate of disappearance of A due to each reaction is assumed to be:

  1. Rate1=K1[A]Rate_1 = K_1 [A]
  2. Rate2=K2[A]Rate_2 = K_2 [A]
  3. Rate3=K3[A]Rate_3 = K_3 [A]

The total rate of disappearance of A is the sum of the rates of the parallel reactions: −d[A]dt=Rate1+Rate2+Rate3- \frac{d[A]}{dt} = Rate_1 + Rate_2 + Rate_3

−d[A]dt=K1[A]+K2[A]+K3[A]- \frac{d[A]}{dt} = K_1 [A] + K_2 [A] + K_3 [A]

−d[A]dt=(K1+K2+K3)[A]- \frac{d[A]}{dt} = (K_1 + K_2 + K_3) [A]

This is a first-order rate law with an effective rate constant Keff=K1+K2+K3K_{eff} = K_1 + K_2 + K_3.

For a first-order reaction, the average lifetime (τ\tau) is the reciprocal of the rate constant.

τ=1Keff\tau = \frac{1}{K_{eff}}

Substitute the given values of the rate constants:

K1=5×10−2s−1K_1 = 5 \times 10^{-2} s^{-1}

K2=3×10−2s−1K_2 = 3 \times 10^{-2} s^{-1}

K3=5×10−3s−1=0.5×10−2s−1K_3 = 5 \times 10^{-3} s^{-1} = 0.5 \times 10^{-2} s^{-1}

Keff=(5×10−2)+(3×10−2)+(0.5×10−2)s−1K_{eff} = (5 \times 10^{-2}) + (3 \times 10^{-2}) + (0.5 \times 10^{-2}) s^{-1}

Keff=(5+3+0.5)×10−2s−1K_{eff} = (5 + 3 + 0.5) \times 10^{-2} s^{-1}

Keff=8.5×10−2s−1K_{eff} = 8.5 \times 10^{-2} s^{-1}

Now calculate the average lifetime:

τ=18.5×10−2s\tau = \frac{1}{8.5 \times 10^{-2}} s

τ=18.5/100s\tau = \frac{1}{8.5/100} s

τ=1008.5s\tau = \frac{100}{8.5} s

To remove the decimal, multiply the numerator and denominator by 10:

τ=100085s\tau = \frac{1000}{85} s