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Chemical Kinetics - Integrated Rate Equations (Zero and First Order)

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Integrated rate equations relate the concentration of reactants with time directly, allowing for the determination of rate constants and half-lives.

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For a Zero-Order Reaction, the rate of reaction is independent of the concentration of reactants. The rate law is expressed as Rate=k[R]0Rate = k[R]^0.

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A plot of [R][R] vs tt for a zero-order reaction gives a straight line with slope=−kslope = -k and intercept=[R]0intercept = [R]_0.

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For a First-Order Reaction, the rate of reaction is proportional to the first power of the concentration of the reactant. The rate law is Rate=k[R]1Rate = k[R]^1.

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A plot of log⁡[R]0[R]\log \frac{[R]_0}{[R]} vs tt for a first-order reaction yields a straight line with slope=k2.303slope = \frac{k}{2.303}.

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The Half-life (t1/2t_{1/2}) of a reaction is the time in which the concentration of a reactant is reduced to one-half of its initial concentration.

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For zero-order reactions, t1/2t_{1/2} is directly proportional to the initial concentration [R]0[R]_0, whereas for first-order reactions, t1/2t_{1/2} is independent of [R]0[R]_0.

📐Formulae

[R]=−kt+[R]0 (Zero Order Integrated Rate Equation)[R] = -kt + [R]_0 \text{ (Zero Order Integrated Rate Equation)}

t1/2=[R]02k (Half-life for Zero Order)t_{1/2} = \frac{[R]_0}{2k} \text{ (Half-life for Zero Order)}

k=2.303tlog⁡[R]0[R] (First Order Integrated Rate Equation)k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \text{ (First Order Integrated Rate Equation)}

t1/2=0.693k (Half-life for First Order)t_{1/2} = \frac{0.693}{k} \text{ (Half-life for First Order)}

[R]=[R]0e−kt (Exponential form for First Order)[R] = [R]_0 e^{-kt} \text{ (Exponential form for First Order)}

💡Examples

Problem 1:

A first-order reaction has a rate constant k=1.15×10−3 s−1k = 1.15 \times 10^{-3} \text{ s}^{-1}. How long will 5 g5 \text{ g} of this reactant take to reduce to 3 g3 \text{ g}?

Solution:

Given: [R]0=5 g[R]_0 = 5 \text{ g}, [R]=3 g[R] = 3 \text{ g}, k=1.15×10−3 s−1k = 1.15 \times 10^{-3} \text{ s}^{-1}. Using the first-order equation: t=2.303klog⁡[R]0[R]t = \frac{2.303}{k} \log \frac{[R]_0}{[R]} t=2.3031.15×10−3log⁡53t = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3} t=2.3031.15×10−3(0.2218)t = \frac{2.303}{1.15 \times 10^{-3}} (0.2218) t≈444 st \approx 444 \text{ s}

Explanation:

We apply the integrated rate law for first-order kinetics. Since the ratio of concentrations is used in the log term, the units of mass (grams) cancel out, allowing us to calculate the time directly.

Problem 2:

The half-life for radioactive decay of 14C^{14}C is 5730 years5730 \text{ years}. An archaeological artifact containing wood had only 80%80\% of the 14C^{14}C found in a living tree. Estimate the age of the sample.

Solution:

Radioactive decay follows first-order kinetics. First, find kk: k=0.693t1/2=0.6935730 year−1k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \text{ year}^{-1} Now, find tt where [R]=0.80[R]0[R] = 0.80 [R]_0: t=2.303klog⁡[R]00.80[R]0t = \frac{2.303}{k} \log \frac{[R]_0}{0.80[R]_0} t=2.303×57300.693log⁡(1.25)t = \frac{2.303 \times 5730}{0.693} \log(1.25) t=19035×0.0969≈1845 yearst = 19035 \times 0.0969 \approx 1845 \text{ years}

Explanation:

All radioactive disintegration reactions follow first-order kinetics. We first determine the decay constant (kk) using the half-life and then use the integrated rate equation to find the age.

Integrated Rate Equations (Zero and First Order) Revision - Class 12 Chemistry CBSE