Understanding the Problem

The reaction is A + B → C + D with the rate law: rate = k [A]^+1/2 [B]^1/2. Note that the exponent for [A] is positive 1/2, not negative. We start with 1 mol each of A and B, and we need to find the time for [A] to become 0.25 mol. The rate constant k is 2.31 × 10^–3 sec^–1.

Step 1: Analyze the Rate Law

The rate law is: rate = k [A]^1/2 [B]^1/2. Since the stoichiometry is 1:1 for A and B, and they start with the same initial concentration, [A] = [B] at all times. Let [A] = [B] = C.

So, the rate law becomes: rate = k (C)^1/2 (C)^1/2 = k C.

Since rate = -d[A]/dt = -dC/dt, we can write:

$-\frac{dC}{dt}=kC$

Step 2: Set Up the Differential Equation

Rearrange the equation:

$\frac{dC}{C}=-kdt$

Step 3: Integrate the Equation

Integrate both sides from initial concentration C₀ to final concentration C_f and from time 0 to t:

${\int }_{C_0}^{C_f}\frac{dC}{C}={\int }_0^t-kdt$

This gives:

$\ln \left(\frac{C_f}{C_0}\right)=-kt$

Step 4: Plug in the Values

Initial concentration C₀ = 1 mol, final concentration C_f = 0.25 mol, k = 2.31 × 10^–3 sec^–1.

$\ln \left(\frac{0.25}{1}\right)=-(2.31\times {10}^{-3})t$

$\ln \left(0.25\right)=-(2.31\times {10}^{-3})t$

ln(0.25) = -1.38629436112

So,

$-1.38629436112=-(2.31\times {10}^{-3})t$

Step 5: Solve for Time t

Multiply both sides by -1:

$1.38629436112=(2.31\times {10}^{-3})t$

Therefore,

$t=\frac{1.38629436112}{2.31\times {10}^{-3}}$

$t=\frac{1.38629436112}{0.00231}\approx 600\text{sec}$

Final Answer

The time taken for the amount of A to become 0.25 mol is 600 sec.

Related Topics

• Rate Laws and Reaction Order
• First-Order Kinetics
• Integrated Rate Laws
• Half-Life of a Reaction

Important Formulae

For a first-order reaction (rate = k [A]):

$\ln \left(\frac{\left[A\right]}{[A{]}_0}\right)=-kt$

Half-life: $t_{1/2}=\frac{\ln \left(2\right)}{k}$

In this specific problem, because [A] = [B], the effective rate law became first-order, allowing us to use the first-order integrated rate law.