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Chapter 14: Problem 17

The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

Short Answer

Expert verified
The percentage of a sample of A that remains unreacted 900 s after a reaction has been started is 10%. The rate of the reaction when [A]=0.50 M is 0.001925 M s^{-1}.

Step by step solution

01

Calculate the Rate Constant from the Half-life

The relationship between the half-life (\(t_{1/2}\)) and the rate constant (k) for a first-order reaction is given by the equation \(t_{1/2} = \frac{0.693}{k}\). We can rearrange this equation to solve for k: \(k = \frac{0.693}{t_{1/2}}\). Substituting the given half-life of 180 s into this equation gives: \(k = \frac{0.693}{180 s} = 0.00385 s^{-1}\).
02

Determine the Percentage of A that Remains Unreacted After 900 s

The amount of reactant A remaining after a certain time (t) in a first-order reaction can be found using the equation \(A_t = A_0 \times e^{-kt}\), where \(A_t\) is the amount of A at time t, \(A_0\) is the initial amount of A, k is the rate constant and t is the time. However, we want to find the percentage of A that remains, so we can normalize this equation to \( \frac{A_t}{A_0} = e^{-kt}\). Substituting the calculated k = 0.00385 s^{-1} and t = 900 s into this equation gives: \( \frac{A_t}{A_0} = e^{-0.00385 s^{-1} \times 900 s}\). This equation gives us approximately 0.10, or 10%. So, 10% of the sample of A remains unreacted after 900 s.
03

Calculate the Rate of Reaction when [A]=0.50 M

The rate of a first-order reaction can be found using the equation \( \text{rate} = k[A] \), where k is the rate constant, and [A] is the concentration of reactant A. Substituting the calculated k = 0.00385 s^{-1} and [A] = 0.50 M into this equation gives: \(\text{rate} = 0.00385 s^{-1} \times 0.50 M = 0.001925 M s^{-1}\). So, the rate of the reaction when [A] = 0.50 M is 0.001925 M s^{-1}.

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Most popular questions from this chapter

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

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.

At \(65^{\circ} \mathrm{C}\), the half-life for the first-order decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})\) is 2.38min. $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ If \(1.00 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is introduced into an evacuated \(15 \mathrm{L}\) flask at \(65^{\circ} \mathrm{C}\) (a) What is the initial partial pressure, in \(\mathrm{mmHg}\), of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) ?\) (b) What is the partial pressure, in \(\mathrm{mmHg}\), of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})\) after \(2.38 \mathrm{min} ?\) (c) What is the total gas pressure, in \(\mathrm{mm} \mathrm{Hg}\), after \(2.38 \mathrm{min} ?\)

In the first-order decomposition of substance A the following concentrations are found at the indicated times: \(t=0 \mathrm{s},[\mathrm{A}]=0.88 \mathrm{M} ; t=50 \mathrm{s},[\mathrm{A}]=0.62 \mathrm{M} ; t=100 \mathrm{s},[\mathrm{A}]=0.44 \mathrm{M} ; t=150 \mathrm{s},[\mathrm{A}]=0.31 \mathrm{M}.\) Calculate the instantaneous rate of decomposition at \(t=100 \mathrm{s}.\)

If the plot of the reactant concentration versus time is nonlinear, but the concentration drops by \(50 \%\) every 10 seconds, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.
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