Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in $0.1 \mathrm{M} \mathrm{HCl}$ occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) $$ \text { is } 8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} $$ (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after $4.00 \times 10^{3} \mathrm{~s}\( if the starting concentration is \)0.500 \mathrm{M}$ ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

Step by step solution

01

Write the rate law for the reaction

The decomposition of urea follows first-order kinetics. Since the reaction is first order in urea and first order overall, the rate law for this reaction is: $$ \text{rate} = k[\mathrm{NH}_{2}\mathrm{CONH}_{2}] $$

02

Calculate the rate constant k using the given rate and initial concentration

We are given that the reaction rate at an initial concentration of \(0.200 M\) is \(8.56 \times 10^{-5} M/s\). We can use this information to find the rate constant k. $$ \text{rate} = k[\mathrm{NH}_{2}\mathrm{CONH}_{2}] $$ Plug in the given values: $$ 8.56 \times 10^{-5} M/s = k (0.200 M) $$ Solve for k: $$ k = \frac{8.56 \times 10^{-5} M/s}{0.200 M} = 4.28 \times 10^{-4} s^{-1} $$

03

Calculate the concentration of urea after a certain time

To find the concentration of urea after \(4.00 \times 10^3 s\), we can use the first-order kinetics equation: $$ \ln\frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t}{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_0} = -kt $$ We know \([\mathrm{NH}_{2}\mathrm{CONH}_{2}]_0 = 0.500 M\), \(k = 4.28 \times 10^{-4} s^{-1}\), and \(t = 4.00 \times 10^3 s\). Now, plug in these values and solve for \([\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t\): $$ \ln\frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t}{0.500 M} = -(4.28 \times 10^{-4} s^{-1})(4.00 \times 10^3 s) $$ $$ \ln\frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t}{0.500 M} = -1.712 $$ Now, find \([\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t\): $$ [\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t = 0.500 M\, e^{-1.712} = 0.213 M $$

04

Calculate the half-life of the reaction

The half-life (\(t_{1/2}\)) of a first-order reaction can be calculated using the formula: $$ t_{1/2} = \frac{0.693}{k} $$ Using the calculated rate constant \(k = 4.28 \times 10^{-4} s^{-1}\), we can find the half-life: $$ t_{1/2} = \frac{0.693}{4.28 \times 10^{-4} s^{-1}} = 1.62 \times 10^3 s $$ In summary: (a) The rate constant k is \(4.28 \times 10^{-4} s^{-1}\). (b) The concentration of urea after \(4.00 \times 10^3 s\) is \(0.213 M\). (c) The half-life for this reaction at \(61.05^{\circ} C\) is \(1.62 \times 10^3 s\).

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