Solveeit Logo
Login

Question

Question: The rate constant for the reaction A→B is 2 × 10⁻⁴ It. mol⁻¹ min⁻¹. The concentration of A at which ...

The rate constant for the reaction A→B is 2 × 10⁻⁴ It. mol⁻¹ min⁻¹. The concentration of A at which rate of the reaction is (1/12) × 10⁻⁵ M sec⁻¹ is :

A

0.25 M

B

(1/20) 5/3\sqrt{5/3} M

C

0.5 M

D

None of these

Answer

0.5 M

Explanation

Solution

The given reaction is A → B. The rate constant for the reaction is k=2×10−4k = 2 \times 10^{-4} L mol⁻¹ min⁻¹. The unit of the rate constant is L mol⁻¹ min⁻¹, which can be written as (mol/L)⁻¹ min⁻¹ or M⁻¹ min⁻¹. For a reaction of order nn, the unit of the rate constant is (concentration)¹⁻ⁿ (time)⁻¹. Comparing the given unit M⁻¹ min⁻¹ with M¹⁻ⁿ min⁻¹, we get 1 - n = -1, which means n = 2. Thus, the reaction is a second-order reaction with respect to A. The rate law for the reaction is given by:

Rate = kk[A]²

We are given the rate of the reaction as Rate = (1/12) × 10⁻⁵ M sec⁻¹. The rate constant is given in min⁻¹, while the rate is given in sec⁻¹. To use the rate law, we need to have consistent time units. Let's convert the rate constant from min⁻¹ to sec⁻¹.

1 minute=60 seconds1 \text{ minute} = 60 \text{ seconds}.

So, 1 min−1=160 sec−11 \text{ min}^{-1} = \frac{1}{60} \text{ sec}^{-1}.

k=2×10−4 M−1 min−1=2×10−4 M−1×160 sec−1k = 2 \times 10^{-4} \text{ M}^{-1} \text{ min}^{-1} = 2 \times 10^{-4} \text{ M}^{-1} \times \frac{1}{60} \text{ sec}^{-1}

k=260×10−4 M−1 sec−1=130×10−4 M−1 sec−1k = \frac{2}{60} \times 10^{-4} \text{ M}^{-1} \text{ sec}^{-1} = \frac{1}{30} \times 10^{-4} \text{ M}^{-1} \text{ sec}^{-1}

k=13×10−5 M−1 sec−1k = \frac{1}{3} \times 10^{-5} \text{ M}^{-1} \text{ sec}^{-1}.

Now, we can use the rate law Rate = kk[A]² and the given rate to find the concentration of A, [A].

(1/12) × 10⁻⁵ M sec⁻¹ = (13×10−5\frac{1}{3} \times 10^{-5} M⁻¹ sec⁻¹) × [A]²

Rearranging the equation to solve for [A]²:

[A]² = (1/12)×10−5 M sec−1(1/3)×10−5 M−1 sec−1\frac{(1/12) \times 10^{-5} \text{ M sec}^{-1}}{(1/3) \times 10^{-5} \text{ M}^{-1} \text{ sec}^{-1}}

[A]² = 1/121/3×10−510−5×M sec−1M−1 sec−1\frac{1/12}{1/3} \times \frac{10^{-5}}{10^{-5}} \times \frac{\text{M sec}^{-1}}{\text{M}^{-1} \text{ sec}^{-1}}

[A]² = 112×3×1×M1−(−1)×sec−1−(−1)\frac{1}{12} \times 3 \times 1 \times \text{M}^{1 - (-1)} \times \text{sec}^{-1 - (-1)}

[A]² = 312×M2×sec0\frac{3}{12} \times \text{M}^2 \times \text{sec}^0

[A]² = 14 M2\frac{1}{4} \text{ M}^2

[A]² = 0.25 M²

Now, take the square root to find [A]:

[A] = 0.25 M2\sqrt{0.25 \text{ M}^2}

[A] = 0.5 M

The concentration of A at which the rate of the reaction is (1/12) × 10⁻⁵ M sec⁻¹ is 0.5 M.