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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 \( t = t_{1⁄2}*\log_2 (1/(1-N)) \). 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

For the disproportionation of \(p\)-toluenesulfinic acid, $$3 \mathrm{ArSO}_{2} \mathrm{H} \longrightarrow \mathrm{ArSO}_{2} \mathrm{SAr}+\mathrm{ArSO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O}$$ (where \(\mathrm{Ar}=p-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4}-\) ), the following data were obtained: \(t=0 \min ,[\mathrm{ArSO}_{2} \mathrm{H}]=0.100 \mathrm{M} ; 15 \mathrm{min}, 0.0863 \mathrm{M} ; 30 \mathrm{min}, 0.0752 \mathrm{M} ; 45 \mathrm{min}, 0.0640 \mathrm{M} ; 60 \mathrm{min}, 0.0568 \mathrm{M} ; 120 \mathrm{min}, 0.0387 \mathrm{M} ; 180 \mathrm{min}, 0.0297 \mathrm{M}; 300 \mathrm{min}, 0.0196 \mathrm{M}.\) (a) Show that this reaction is second order. (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0500 \mathrm{M} ?\) (d) At what time would \(\left(\mathrm{ArSO}_{2} \mathrm{H}\right)=0.0250 \mathrm{M} ?\) (e) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0350 \mathrm{M} ?\)

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One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.

If the plot of the reactant concentration versus time is linear, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.

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?
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