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Chapter 14: Problem 55

The first-order reaction \(A \longrightarrow\) products has a halflife, \(t_{1 / 2},\) of 46.2 min at \(25^{\circ} \mathrm{C}\) and \(2.6 \mathrm{min}\) at \(102^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) At what temperature would the half-life be 10.0 min?

Short Answer

Expert verified
The activation energy of the reaction is found to be approximately \(61.5 \: kJ/mol\). The temperature at which the half-life would be \(10\: min\) is around \(56.4^{\circ}C\).

Step by step solution

01

Write down the Equations and Given Data

First, let’s write down the Arrhenius equation: \(k=Ae^{-E_a / RT}\), where k is the rate constant, A is the frequency factor, Eₐ is the activation energy, R is the gas constant and T is the temperature. Also for a first order reaction, the relation between the half-life and the rate constant is \(t_{1/2}=\frac{0.693}{k}\). Given, half life at \(25^{\circ} \mathrm{C} (298 K)\) is \(46.2 \min\), and at \(102^{\circ} \mathrm{C} (375 K)\) it is \(2.6 \min\). The temperature is in Kelvin, because it is an absolute measurement scale.
02

Calculate Activation Energy

Since the frequency factor 'A' is the same for both temperatures and the T is always changing when measuring half-lives, we can set the two rate constants (k1 and k2) equal to each other and solve for the activation energy (Ea). From the given two temperatures and half-lives, write the Arrhenius equation for each and divide one by the other to eliminate 'A'. This gives \(\frac{k1}{k2} = \frac{e^{-E_{a}/RT_{1}}}{e^{-E_{a}/RT_{2}}}\) which simplifies to: \( \frac{-E_{a}}{RT_{1}} + \frac{E_{a}}{RT_{2}} = ln(\frac{k1}{k2}) \).Substitute the expression of rate constant from the relation \(t_{1/2} = \frac{0.693}{k}\) in the above equation and solve for` Ea` to get the Activation Energy.
03

Calculate the Temperature for a Specific Half-Life

We are supposed to find the temperature at which the half-life is 10.0 min. We can use the same procedure as in step 2, but this time we are solving for T when \(t_{1/2} = 10.0 min\). Writing a similar equation like in Step 2 and substituting the values, we can solve for T. Do remember to convert minutes into seconds to keep the consistancy in units.

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Most popular questions from this chapter

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

Hydroxide ion is involved in the mechanism of the following reaction but is not consumed in the overall reaction. $$\mathrm{OCI}^{-}+\mathrm{I}^{-} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{OI}^{-}+\mathrm{Cl}^{-}$$ (a) From the data given, determine the order of the reaction with respect to \(\mathrm{OCl}^{-}, \mathrm{I}^{-},\) and \(\mathrm{OH}^{-}\) (b) What is the overall reaction order? (c) Write the rate equation, and determine the value of the rate constant, \(k.\) $$\begin{array}{lccl} \hline & & & \text { Rate Formation } \\ {\left[\mathrm{OC}^{-}\right], \mathrm{M}} & {\left[\mathrm{l}^{-}\right], \mathrm{M}} & {\left[\mathrm{OH}^{-}\right], \mathrm{M}} & \mathrm{O}^{-}, \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.0040 & 0.0020 & 1.00 & 4.8 \times 10^{-4} \\ 0.0020 & 0.0040 & 1.00 & 5.0 \times 10^{-4} \\ 0.0020 & 0.0020 & 1.00 & 2.4 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.50 & 4.6 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.25 & 9.4 \times 10^{-4} \\ \hline \end{array}$$

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.

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

In the first-order reaction \(A \longrightarrow\) products, \([\mathrm{A}]=0.816 \mathrm{M}\) initially and \(0.632 \mathrm{M}\) after \(16.0 \mathrm{min}.\) (a) What is the value of the rate constant, \(k ?\) (b) What is the half-life of this reaction? (c) At what time will \([\mathrm{A}]=0.235 \mathrm{M} ?\) (d) What will [A] be after 2.5 h?
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