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. (b) Assuming the collision factor is the same for both situations, estimate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

The enzyme urease catalyzes the reaction of urea with water to produce carbon dioxide and ammonia. The balanced equation for this reaction is: \( \mathrm{NH}_2\mathrm{CONH}_2 + H_2O \xrightarrow{urease} CO_2 + 2NH_3 \)

We can use the Arrhenius equation to estimate the difference in activation energies for the uncatalyzed and enzyme-catalyzed reactions. The Arrhenius equation is: \( k = A \times e^{\frac{-E_A}{RT}} \) where \(k\) is the rate constant, \(A\) is the Arrhenius constant (also known as the collision factor), \(E_A\) is the activation energy, \(R\) is the gas constant (\(\approx 8.314 \text{ J mol}^{-1} \text{ K}^{-1}\)), and \(T\) is the temperature in Kelvin. We are given that the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \text{ s}^{-1}\) without the enzyme at \(100^{\circ} \text{C}\), and with a rate constant of \(3.4 \times 10^{4} \text{ s}^{-1}\) in the presence of the enzyme at \(21^{\circ} \text{C}\). Since the problem states that the collision factor is the same for both situations, we specifically can write: \( \frac{k_{uncatalyzed}}{k_{catalyzed}} = \frac{A \times e^{\frac{-E_A^{uncatalyzed}}{RT_{uncatalyzed}}}}{A \times e^{\frac{-E_A^{catalyzed}}{RT_{catalyzed}}}} \) Where: - \(k_{uncatalyzed} = 4.15 \times 10^{-5} \text{ s}^{-1}\) - \(k_{catalyzed} = 3.4 \times 10^{4} \text{ s}^{-1}\) - \(T_{uncatalyzed} = 373.15 \text{ K}\) (converting from Celsius to Kelvin) - \(T_{catalyzed} = 294.15 \text{ K }\) (converting from Celsius to Kelvin) Now we need to find the difference in activation energies, \(\Delta E_A = E_A^{uncatalyzed} - E_A^{catalyzed}\), using the given information. Since the collision factors (A) in the numerator and denominator are the same, they cancel out: \( \frac{k_{uncatalyzed}}{k_{catalyzed}} = \frac{e^{\frac{-E_A^{uncatalyzed}}{RT_{uncatalyzed}}}}{e^{\frac{-E_A^{catalyzed}}{RT_{catalyzed}}}} \) Taking the natural logarithm of both sides: \( \ln{\frac{k_{uncatalyzed}}{k_{catalyzed}}} = \frac{-E_A^{uncatalyzed}}{RT_{uncatalyzed}} + \frac{E_A^{catalyzed}}{RT_{catalyzed}} \) Rearranging the equation for \(\Delta E_A\): \( \Delta E_A = \left( \frac{-RT_{catalyzed}}{RT_{uncatalyzed}} + 1 \right)R \ln{\frac{k_{uncatalyzed}}{k_{catalyzed}}} \) Plugging in the given values and solving for \(\Delta E_A\): \( \Delta E_A = \left( \frac{-294.15}{373.15} + 1 \right)(8.314) \ln{\frac{4.15 \times 10^{-5}}{3.4 \times 10^{4}}} \) \( \Delta E_A \approx 51.4 \text{ kJ mol}^{-1} \) The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately 51.4 kJ/mol.