The enzyme urease catalyzes the reaction of urea, $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$, with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at $100^{\circ} \mathrm{C}$. In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at $21^{\circ} \mathrm{C}$. (a) Write out the balanced equation for the reaction catalyzed by urease. \((\mathbf{b})\) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

The balanced equation for the reaction of urea with water in the presence of urease enzyme is: \[NH_2CONH_2 + H_2O \xrightarrow{Urease} 2 NH_3 + CO_2\]

First, note that the rate constant of the uncatalyzed reaction is given at 100°C and the rate constant of the catalyzed reaction is given at 21°C. We are asked to assume that the rate of the catalyzed reaction is the same at 100°C as it is at 21°C. Using the Arrhenius equation: \(k = A\mathrm{e}^{-\frac{Ea}{RT}}\) Where: - k is the rate constant - A is the pre-exponential factor - Ea is the activation energy - R is the gas constant (8.314 J/mol K) - T is the temperature in Kelvin (K) First, we need to convert the temperatures to Kelvin: \(100^{\circ}C + 273.15 = 373.15K\) \(21^{\circ}C + 273.15 = 294.3K\) The difference in activation energy (ΔEa) can be calculated as follows: \(ΔEa = Ea_{uncatalyzed} - Ea_{catalyzed}\) To find the activation energies (Ea) for the uncatalyzed and catalyzed reactions, we can use the Arrhenius equation and solve for Ea: \(Ea_{uncatalyzed} = -R\ln(\frac{k_{uncatalyzed}}{A})\) \(Ea_{catalyzed} = -R\ln(\frac{k_{catalyzed}}{A})\) Where A can be assumed to be equal for both reactions. We can then find the difference in activation energy as: \(ΔEa = -R\ln(\frac{k_{uncatalyzed}}{A}) + R\ln(\frac{k_{catalyzed}}{A})\) \(ΔEa = R\ln(\frac{k_{catalyzed}}{k_{uncatalyzed}})\) Next, plug in the given values for the rate constants and the gas constant to find ΔEa: \(ΔEa = (8.314 \frac{J}{molK})\ln(\frac{3.4 \times 10^{4} s^{-1}}{4.15 \times 10^{-5} s^{-1}})\) \(ΔEa ≈ 186.3 kJ/mol\)

In actuality, enzymes typically lose their effectiveness at such high temperatures (100°C) due to denaturation. This means that the rate of the catalyzed reaction at 100°C would likely be significantly decreased compared to that at 21°C.

The difference in activation energy between the catalyzed and uncatalyzed reactions is approximately 186.3 kJ/mol, assuming the rate of the catalyzed reaction is the same at 100°C as it is at 21°C. However, considering that enzymes denature at high temperatures, the actual difference in activation energies would be smaller because the catalyzed reaction rate would be slower at 100°C.