JEE Challenger
More from Chemical Kinetics
ChemMedAdv 2021 • Paper 2

Rate Constant and Reaction Rates in First Order Kinetics

For the given chemical reaction: 2X+Y→kP2\mathbf{X} + \mathbf{Y} \xrightarrow{k} \mathbf{P} the rate of reaction is given by d[P]dt=k[X]\frac{d[\mathbf{P}]}{dt} = k[\mathbf{X}]. Two moles of X\mathbf{X} are mixed with one mole of Y\mathbf{Y} to yield 1.0 L1.0\text{ L} of solution. At 50 s50\text{ s}, 0.5 mole0.5\text{ mole} of Y\mathbf{Y} remains in the reaction mixture. Select the correct statement(s) regarding this reaction.

(Given: ln⁡2=0.693\ln 2 = 0.693)

Options

A

The rate constant, kk, of the reaction is 13.86×10−4 s−113.86 \times 10^{-4}\text{ s}^{-1}.

B

Half-life of X\mathbf{X} is 50 s50\text{ s}.

Correct
C

At 50 s50\text{ s}, −d[X]dt=13.86×10−3 mol L−1 s−1-\frac{d[\mathbf{X}]}{dt} = 13.86 \times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}.

Correct
D

At 100 s100\text{ s}, −d[Y]dt=3.46×10−3 mol L−1 s−1-\frac{d[\mathbf{Y}]}{dt} = 3.46 \times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}.

Correct

Topics & Concepts

Chemical KineticsRate of ReactionOrder of Reaction

Step-by-Step Solution

For the chemical reaction: 2X+Y→kP2\mathbf{X} + \mathbf{Y} \xrightarrow{k} \mathbf{P}

The differential rate equation is given as: d[P]dt=k[X]\frac{d[\mathbf{P}]}{dt} = k[\mathbf{X}]

Using stoichiometric coefficients, the rates of reaction with respect to reactants and products are related as: Rate of reaction=−12d[X]dt=−d[Y]dt=d[P]dt=k[X]\text{Rate of reaction} = -\frac{1}{2}\frac{d[\mathbf{X}]}{dt} = -\frac{d[\mathbf{Y}]}{dt} = \frac{d[\mathbf{P}]}{dt} = k[\mathbf{X}]

From this, we get: −d[X]dt=2k[X]-\frac{d[\mathbf{X}]}{dt} = 2k[\mathbf{X}] −d[Y]dt=k[X]-\frac{d[\mathbf{Y}]}{dt} = k[\mathbf{X}]

1. Initial Concentrations and Concentration at t=50 st = 50\text{ s}

Given initial moles in a volume of 1.0 L1.0\text{ L}: [X]0=2.0 mol1.0 L=2.0 M[\mathbf{X}]_0 = \frac{2.0\text{ mol}}{1.0\text{ L}} = 2.0\text{ M} [Y]0=1.0 mol1.0 L=1.0 M[\mathbf{Y}]_0 = \frac{1.0\text{ mol}}{1.0\text{ L}} = 1.0\text{ M}

At t=50 st = 50\text{ s}, the remaining moles of Y\mathbf{Y} is 0.5 mol0.5\text{ mol}: [Y]50=0.5 M[\mathbf{Y}]_{50} = 0.5\text{ M}

The amount of Y\mathbf{Y} consumed in 50 s50\text{ s}: Δ[Y]=[Y]0−[Y]50=1.0−0.5=0.5 M\Delta [\mathbf{Y}] = [\mathbf{Y}]_0 - [\mathbf{Y}]_{50} = 1.0 - 0.5 = 0.5\text{ M}

Since 2 moles2\text{ moles} of X\mathbf{X} react for every 1 mole1\text{ mole} of Y\mathbf{Y}: Amount of X consumed in 50 s=2×0.5 M=1.0 M\text{Amount of }\mathbf{X}\text{ consumed in } 50\text{ s} = 2 \times 0.5\text{ M} = 1.0\text{ M}

Thus, the remaining concentration of X\mathbf{X} at t=50 st = 50\text{ s} is: [X]50=[X]0−1.0=2.0−1.0=1.0 M[\mathbf{X}]_{50} = [\mathbf{X}]_0 - 1.0 = 2.0 - 1.0 = 1.0\text{ M}

2. Half-Life of X\mathbf{X} (Checking Option B)

Since [X][\mathbf{X}] reduces from 2.0 M2.0\text{ M} to 1.0 M1.0\text{ M} (half of its initial value) in 50 s50\text{ s}, and the reaction is first-order with respect to X\mathbf{X}, the half-life of X\mathbf{X} is: t1/2(X)=50 st_{1/2}(\mathbf{X}) = 50\text{ s}

Thus, Option B is CORRECT.

3. Rate Constant kk (Checking Option A)

For first-order kinetics with respect to X\mathbf{X}: −d[X]dt=2k[X]-\frac{d[\mathbf{X}]}{dt} = 2k[\mathbf{X}]

Integrating gives: ln⁡([X]0[X]t)=2kt\ln\left(\frac{[\mathbf{X}]_0}{[\mathbf{X}]_t}\right) = 2kt

Using t1/2=50 st_{1/2} = 50\text{ s}: 2k=ln⁡2t1/2=0.69350 s=1.386×10−2 s−12k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{50\text{ s}} = 1.386 \times 10^{-2}\text{ s}^{-1} k=1.386×10−22=6.93×10−3 s−1=69.3×10−4 s−1k = \frac{1.386 \times 10^{-2}}{2} = 6.93 \times 10^{-3}\text{ s}^{-1} = 69.3 \times 10^{-4}\text{ s}^{-1}

Option A claims k=13.86×10−4 s−1k = 13.86 \times 10^{-4}\text{ s}^{-1}, which is incorrect. Thus, Option A is INCORRECT.

4. Rate of Disappearance of X\mathbf{X} at t=50 st = 50\text{ s} (Checking Option C)

At t=50 st = 50\text{ s}, [X]=1.0 M[\mathbf{X}] = 1.0\text{ M}. −d[X]dt=2k[X]=(1.386×10−2 s−1)×(1.0 mol L−1)=13.86×10−3 mol L−1 s−1-\frac{d[\mathbf{X}]}{dt} = 2k[\mathbf{X}] = (1.386 \times 10^{-2}\text{ s}^{-1}) \times (1.0\text{ mol L}^{-1}) = 13.86 \times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}

Thus, Option C is CORRECT.

5. Rate of Disappearance of Y\mathbf{Y} at t=100 st = 100\text{ s} (Checking Option D)

Since 100 s100\text{ s} corresponds to two half-lives (2×t1/22 \times t_{1/2}): [X]100=[X]022=2.0 M4=0.5 M[\mathbf{X}]_{100} = \frac{[\mathbf{X}]_0}{2^2} = \frac{2.0\text{ M}}{4} = 0.5\text{ M}

The rate of disappearance of Y\mathbf{Y} at t=100 st = 100\text{ s} is: −d[Y]dt=k[X]=(6.93×10−3 s−1)×(0.5 mol L−1)=3.465×10−3 mol L−1 s−1≈3.46×10−3 mol L−1 s−1-\frac{d[\mathbf{Y}]}{dt} = k[\mathbf{X}] = (6.93 \times 10^{-3}\text{ s}^{-1}) \times (0.5\text{ mol L}^{-1}) = 3.465 \times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1} \approx 3.46 \times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}

Thus, Option D is CORRECT.

Conclusion

The correct options are B, C, and D.

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