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Ch.14 - Chemical Kinetics
Chapter 14, Problem 109

A certain substance X decomposes. Fifty percent of X remains after 100 minutes. How much X remains after 200 minutes if the reaction order with respect to X is (c) second order?

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1
Identify the reaction order: The problem states that the reaction is second order with respect to X.
Use the second-order integrated rate law: \( \frac{1}{[X]} = kt + \frac{1}{[X]_0} \), where \([X]_0\) is the initial concentration, \([X]\) is the concentration at time \(t\), and \(k\) is the rate constant.
Determine the rate constant \(k\): Use the information that 50% of X remains after 100 minutes. This means \([X] = \frac{1}{2}[X]_0\) at \(t = 100\) minutes. Substitute these values into the second-order rate law to solve for \(k\).
Calculate \([X]\) at 200 minutes: Use the value of \(k\) obtained from the previous step and substitute \(t = 200\) minutes into the second-order rate law to find \([X]\) at this time.
Interpret the result: The calculated \([X]\) will tell you how much of the substance X remains after 200 minutes.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Reaction Order

Reaction order refers to the power to which the concentration of a reactant is raised in the rate law of a chemical reaction. It indicates how the rate of reaction depends on the concentration of reactants. For example, a second-order reaction means that the rate is proportional to the square of the concentration of the reactant, which significantly influences how the concentration changes over time.
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Half-Life in Second-Order Reactions

The half-life of a substance in a second-order reaction is inversely proportional to the initial concentration of the reactant. Unlike first-order reactions, where the half-life is constant, the half-life for second-order reactions increases as the reaction proceeds. This means that as time goes on, it takes longer for the concentration of the reactant to decrease by half.
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Integrated Rate Law for Second-Order Reactions

The integrated rate law for a second-order reaction can be expressed as 1/[X] = kt + 1/[X]₀, where [X] is the concentration of the reactant at time t, k is the rate constant, and [X]₀ is the initial concentration. This equation allows us to calculate the concentration of the reactant at any given time, which is essential for determining how much of substance X remains after a specified duration.
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Related Practice
Textbook Question

Consider the reaction: 2 NH3(aq) + OCl-(aq) → N2H4(aq) + H2O(l) + Cl- (aq) This three-step mechanism is proposed: NH3(aq) + OCl- (aq) Δk1k2 NH2Cl(aq) + OH- (aq) Fast NH2Cl(aq) + NH3(aq) →k3 N2H5+ (aq) + Cl- (aq) Slow N2H5+ (aq) + OH-(aq) →k4 N2H4(aq) + H2O(l) Fast a. Show that the mechanism sums to the overall reaction.

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Open Question
What rate law corresponds to the proposed mechanism for the formation of hydrogen bromide, which can be written in a simplified form as: Br2(g) → 2Br(g) (Fast) Br(g) + H2(g) → HBr(g) + H(g) (Slow) H(g) + Br2(g) → HBr(g) + Br(g) (Fast)?
Open Question
What rate law corresponds to the proposed mechanism for the formation of hydrogen iodide, which can be written in simplified form as: I2 Δk1k-1 2I (Fast), I + H2 Δk2k-2 H2I (Fast), H2I + I ¡k3 2HI (Slow)?
Textbook Question

The half-life for radioactive decay (a first-order process) of plutonium- 239 is 24,000 years. How many years does it take for one mole of this radioactive material to decay until just one atom remains?

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Open Question
The energy of activation for the decomposition of 2 mol of HI to H2 and I2 in the gas phase is 185 kJ. The heat of formation of HI(g) from H2(g) and I2(g) is -5.65 kJ/mol. Find the energy of activation for the reaction of 1 mol of H2 and 1 mol of I2 to form 2 mol of HI in the gas phase.
Open Question
Ethyl chloride vapor decomposes by the first-order reaction: C2H5Cl -> C2H4 + HCl. The activation energy is 249 kJ/mol, and the frequency factor is 1.6 * 10^14 s^-1. Find the temperature at which the rate of the reaction would be twice as fast.