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 \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 M,\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\) , (a) What is the rate constant, \(k ?\) units of \(s^{-1}\) . (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to 0.100\(?\)

Step by step solution

01

Calculate the rate constant, k

Since the given reaction is first-order overall and first-order with respect to urea, the rate expression can be written as: Rate = \(k[\mathrm{NH}_{2}\mathrm{CONH}_{2}]\) Given that the initial concentration of urea (\([\mathrm{NH}_{2}\mathrm{CONH}_{2}]\)) is 0.200 M and the rate at this concentration is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). We can now calculate the rate constant \(k\) using the above equation: \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} = k(0.200 \mathrm{M})\) Rearrange to find the value of \(k\): \(k = \frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}}\)

02

Calculate the half-life of the reaction

For a first-order reaction, the half-life (\(t_{1/2}\)) can be calculated using the following equation: \(t_{1/2} = \frac{0.693}{k}\) Now, we plug the value of rate constant \(k\) we calculated in Step 1 into the formula: \(t_{1/2} = \frac{0.693}{\frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}}}\)

03

Calculate the time required for the absorbance to fall to 0.100

To find the time required for absorbance to fall to 0.100, we can use the following first-order integrated rate law equation: \(ln \frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_{initial}}{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_{final}} = kt\) We know the initial concentration and the absorbance we want to achieve, as well as the rate constant \(k\) from Step 1. So, let's plug it in to find the time, \(t\): \(ln \frac{0.200 \mathrm{M}}{0.100 \mathrm{M}} = \left(\frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}}\right)t\) Rearrange and solve for \(t\) to find the time required for the absorbance to fall to 0.100.

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