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Chapter 14: Problem 78

A certain first-order reaction is 35.5 percent complete in 4.90 min at \(25^{\circ} \mathrm{C}\). What is its rate constant?

Short Answer

Expert verified
The rate constant (k) is obtained by solving the equation with the given values, which is generally in the units of 1/s for first-order reactions. Remember to change the time to seconds for meaningful comparison and standard analysis in chemistry.

Step by step solution

01

Identifying given values

First, identify the known values from the given problem. Here, it is given that the reaction is 35.5 percent complete, which in fraction amounts to 0.355 (X = 0.355). The time it takes for the reaction to reach this point is 4.90 minutes (t = 4.90).
02

Converting time to appropriate unit

It is generally accepted to have the rate constant k in units of 1/s for first-order reactions. So, convert the given time from minutes to seconds. That is, \( t = 4.90 \times 60 = 294 \) seconds.
03

Solving for rate constant (k)

Substitute these values back into the equation \( k = \frac{-\ln(1-X)}{t} \). You will get \( k = \frac{-\ln(1-0.355)}{294} \). Solving the above equation will yield a numerical value for k.

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Most popular questions from this chapter

What is meant by the rate of a chemical reaction?

Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons_{-1}} \mathrm{ES} \quad \text { (fast equilbrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)

Consider the reaction $$ \mathrm{A} \longrightarrow \mathrm{B} $$ The rate of the reaction is \(1.6 \times 10^{-2} M / \mathrm{s}\) when the concentration of A is \(0.35 M\). Calculate the rate constant if the reaction is (a) first order in \(\mathrm{A},\) (b) second order in A.

The rate constants of some reactions double with every 10 -degree rise in temperature. Assume a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

The equation for the combustion of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \longrightarrow 4 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ Explain why it is unlikely that this equation also represents the elementary step for the reaction.
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