Question

Suppose that a certain biologically important reaction is quite slow at physiological temperature $\left({37}^{\circ }\mathrm{C}\right)$ in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a $1\times {10}^5$ -fold increase in the reaction rate?

Short answer

To achieve a $1\times {10}^5$-fold increase in the reaction rate, the enzyme must lower the activation energy of the reaction by approximately 169.6 kJ/mol.

Step-by-step solution

Write down the Arrhenius equation

The Arrhenius equation is given by: $k=A{e}^{-\frac{E_a}{RT}}$ where: - $k$ is the rate constant of the reaction - $A$ is the pre-exponential factor (also known as the collision factor) - $E_a$ is the activation energy of the reaction - $R$ is the gas constant, approximately equal to 8.314 J/(mol*K) - $T$ is the temperature in Kelvin

Convert the temperature to Kelvin

The given temperature is 37°C. To convert it to Kelvin, we add 273.15: $T=37+273.15=310.15\mathrm{K}$

Set up the ratio of rate constants

We want to find the reduction in activation energy needed for a 100,000-fold increase in the reaction rate. We will call the initial activation energy $E_{a1}$ and the modified activation energy (with the enzyme) $E_{a2}$. Then, the desired ratio of rate constants is: $\frac{k_2}{k_1}=1\times {10}^5$ Using the Arrhenius equation, we can write: $\frac{A{e}^{-\frac{E_{a2}}{RT}}}{A{e}^{-\frac{E_{a1}}{RT}}}=1\times {10}^5$

Simplify and solve for delta E

Since the collision factor (pre-exponential factor) A remains the same, we can simplify the equation: $e^{\frac{E_{a1}-E_{a2}}{RT}}=1\times {10}^5$ Now we will solve for the difference in activation energies $\mathrm{\Delta }E=E_{a1}-E_{a2}$: $\mathrm{\Delta }E=RT\cdot \ln (1\times {10}^5)$ Plugging in the values for R and T: $\mathrm{\Delta }E=(8.314\frac{\mathrm{J}}{\mathrm{mol}\cdot \mathrm{K}})(310.15\mathrm{K})\ln (1\times {10}^5)$ Calculating the result: $\mathrm{\Delta }E\approx 1.696\times {10}^5\mathrm{J}/\mathrm{mol}$

Express the result in kJ/mol

To express the result in kJ/mol, we divide by 1000: $\mathrm{\Delta }E\approx \frac{1.696\times {10}^5}{1000}\mathrm{kJ}/\mathrm{mol}=169.6\mathrm{kJ}/\mathrm{mol}$ So, the enzyme must lower the activation energy of the reaction by approximately 169.6 kJ/mol to achieve a 100,000-fold increase in the reaction rate.