/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A reaction is \(50 \%\) complete... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 34

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

Short Answer

Expert verified
The time after the start at which the reaction will be 75% complete is approximately 44.7 minutes for a first order reaction and approximately 45.0 minutes for a zero order reaction.

Step by step solution

01

Analyze the given problem

The exercise provides that the reaction is 50% complete in 30.0 minutes and it is required to calculate the time required for the reaction to be 75% complete for a first order and zero order reaction. The amount of the reaction completed is given by \(N = 1 - e^{-kt}\). For a first order reaction, this simplifies to \( t = t_{1鈦�2}*\log_2 (1/(1-N)) \). For a zero order reaction, we have \( N = kt \)
02

Calculate time for a first order reaction

Using given values and first order rate law we have: \( t = (30min)*\log_2(1/(1-0.75)) \). Evaluating this yields approximately 44.7 minutes.
03

Calculate time for a zero order reaction

For zero order reaction, the formula becomes \( N = kt \). Solving for k from existing data we get \( k = N/t = 0.50/30min \). Substituting this k into the equation for N=0.75 we find \( t = N/k = 0.75/(0.50/30min) \). Evaluating this gives approximately 45.0 minutes.

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Most popular questions from this chapter

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

The decomposition of ethylene oxide at \(690 \mathrm{K}\) is monitored by measuring the total gas pressure as a function of time. The data obtained are \(t=10 \mathrm{min}, P_{\text {tot }}= 139.14 \mathrm{mmHg} ; 20 \mathrm{min}, 151.67 \mathrm{mmHg} ; 40 \mathrm{min}, 172.65 \mathrm{mmHg} ; 60 \mathrm{min}, 189.15 \mathrm{mmHg} ;\) \(100 \mathrm{min}, 212.34\) \(\mathrm{mmHg} ; 200 \mathrm{min}, 238.66 \mathrm{mmHg} ; \infty, 249.88 \mathrm{mmHg}\) What is the order of the reaction \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) ?\)

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

Explain why (a) A reaction rate cannot be calculated from the collision frequency alone. (b) The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency increases much more slowly. (c) The addition of a catalyst to a reaction mixture can have such a pronounced effect on the rate of a reaction, even if the temperature is held constant.
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