Question

For the simultaneous reactions:

$2A \rightarrow B; -\frac{d[A]}{dt} = 0.04 \ min^{-1} [A]$

$3A \rightarrow 2C; -\frac{d[A]}{dt} = 0.03 \ min^{-1} [A]$

Half-life for the disappearance of 'A' is x min (In 2 = 0.7)

The mole percent of 'B' in the total product formed up to 20 min from the start of reaction will be y

Question 31:

The value of x is ______.

Answer 1

10

Answer 2

The problem involves simultaneous first-order reactions. We need to calculate the half-life for the disappearance of 'A' (denoted as 'x') and the mole percent of 'B' in the total product (denoted as 'y').

Part 1: Calculate the value of x (Half-life for the disappearance of 'A')

We are given two simultaneous reactions with their respective rate expressions for the disappearance of A:

1. $2A \rightarrow B$; $-\frac{d[A]}{dt}_1 = k_1[A]$ where $k_1 = 0.04 \ min^{-1}$
2. $3A \rightarrow 2C$; $-\frac{d[A]}{dt}_2 = k_2[A]$ where $k_2 = 0.03 \ min^{-1}$

For simultaneous first-order reactions, the overall rate of disappearance of A is the sum of the rates of disappearance through each pathway: $-\frac{d[A]}{dt}_{overall} = -\frac{d[A]}{dt}_1 + -\frac{d[A]}{dt}_2$ $-\frac{d[A]}{dt}_{overall} = k_1[A] + k_2[A]$ $-\frac{d[A]}{dt}_{overall} = (k_1 + k_2)[A]$

The overall rate constant for the disappearance of A, $k_{overall}$, is the sum of the individual rate constants: $k_{overall} = k_1 + k_2 = 0.04 \ min^{-1} + 0.03 \ min^{-1} = 0.07 \ min^{-1}$

For a first-order reaction, the half-life ($t_{1/2}$) is given by the formula: $t_{1/2} = \frac{\ln 2}{k_{overall}}$

Given $\ln 2 = 0.7$: $x = t_{1/2} = \frac{0.7}{0.07 \ min^{-1}} = 10 \ min$

Part 2: Calculate the value of y (Mole percent of 'B' in the total product formed up to 20 min)

To find the mole percent of B in the total product (B + C), we need to determine the relative rates of formation of B and C.

From reaction 1: $2A \rightarrow B$ The rate of formation of B is related to the rate of disappearance of A by the stoichiometric coefficient: $\frac{d[B]}{dt} = -\frac{1}{2}\frac{d[A]}{dt}_1 = \frac{1}{2} k_1[A]$ $\frac{d[B]}{dt} = \frac{1}{2} (0.04 \ min^{-1})[A] = 0.02[A]$

From reaction 2: $3A \rightarrow 2C$ The rate of formation of C is related to the rate of disappearance of A by the stoichiometric coefficient: $\frac{d[C]}{dt} = -\frac{2}{3}\frac{d[A]}{dt}_2 = \frac{2}{3} k_2[A]$ $\frac{d[C]}{dt} = \frac{2}{3} (0.03 \ min^{-1})[A] = 0.02[A]$

Comparing the rates of formation: $\frac{d[B]}{dt} = 0.02[A]$ $\frac{d[C]}{dt} = 0.02[A]$

Since $\frac{d[B]}{dt} = \frac{d[C]}{dt}$, the rate of formation of B is equal to the rate of formation of C. This implies that over any given time period, the moles of B formed ($N_B$) will be equal to the moles of C formed ($N_C$). $N_B = N_C$

The total moles of product formed is $N_{total} = N_B + N_C$. Substituting $N_C = N_B$: $N_{total} = N_B + N_B = 2N_B$

The mole percent of B in the total product is: Mole % of B = $\frac{N_B}{N_{total}} \times 100\%$ Mole % of B = $\frac{N_B}{2N_B} \times 100\% = \frac{1}{2} \times 100\% = 50\%$

So, the value of y is 50. The time (20 min) is irrelevant for this calculation because the ratio of products formed in parallel first-order reactions is constant throughout the reaction.

The question asks for the value of x.

The value of x is 10.