Problem 93 - Chemistry The Central Science

Question

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Step-by-Step Solution

Answer

The difference in activation energies for the uncatalyzed and enzyme-catalyzed reactions is approximately \( -48,300\, J/mol\). The negative sign indicates that the activation energy of the enzyme-catalyzed reaction is lower than that of the uncatalyzed reaction.

Step 1: Recall the Arrhenius Equation

We will use the Arrhenius equation, which describes the temperature dependence of reaction rates: \[k = Ae^{\frac{-E_a}{RT}}\] 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), and T is the temperature in Kelvin.

Step 2: Convert Temperature to Kelvin

We need to convert the given temperature from Celsius to Kelvin. Temperature in Kelvin can be calculated using the formula: \[T(K) = T(°C) + 273.15\] For the given temperature of 25°C, we have: \[T(K) = 25 + 273.15 = 298.15\;K\]

Step 3: Solve for Activation Energies

We first solve for activation energy in both situations (uncatalyzed and enzyme-catalyzed reactions) using the Arrhenius equation. Let \(E_{a1}\) be the activation energy for the uncatalyzed reaction and \(E_{a2}\) be the activation energy for the enzyme-catalyzed reaction: For the uncatalyzed reaction: \[k_1 = Ae^{\frac{-E_{a1}}{RT}}\] For the enzyme-catalyzed reaction: \[k_2 = Ae^{\frac{-E_{a2}}{RT}}\] We are given that the collision factor (A) is the same in both situations. So, we can divide the second equation by the first one: \[\frac{k_2}{k_1} = e^{\frac{E_{a1} - E_{a2}}{RT}}\] Our next objective is to isolate the term \((E_{a1} - E_{a2})\).

Step 4: Isolate the Difference in Activation Energies

To find the difference in activation energies, we need to isolate the term \((E_{a1} - E_{a2})\). We can do this by taking the natural logarithm of both sides of the equation: \[\ln\left(\frac{k_2}{k_1}\right) = \frac{E_{a1} - E_{a2}}{RT}\] Now, we can solve for \((E_{a1} - E_{a2})\): \[(E_{a1} - E_{a2}) = RT\ln\left(\frac{k_2}{k_1}\right)\]

Step 5: Calculate the Difference in Activation Energies

Now, we substitute the given values for \(k_1\), \(k_2\) and the temperature (in Kelvin) into the equation to calculate the difference in activation energies: \[(E_{a1} - E_{a2}) = (8.314\, J/mol\, K)(298.15\, K)\ln\left(\frac{1.0 \times 10^{6}\, s^{-1}}{0.039\, s^{-1}}\right)\] \[(E_{a1} - E_{a2}) \approx -48,300\, J/mol\] The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately -48,300 J/mol. The negative sign indicates that the activation energy of the enzyme-catalyzed reaction is lower than that of the uncatalyzed reaction.

Key Concepts

Arrhenius Equation

When a chemical reaction takes place, the rate at which it proceeds is highly dependent on temperature. The Arrhenius equation gives us a formula to understand this relationship. It is expressed as:\[k = A e^{\frac{-E_a}{RT}}\]In this formula:- \( k \) is the reaction rate constant, which tells us how quickly a reaction proceeds.- \( A \) is the pre-exponential factor, a constant that is characteristic of each chemical reaction.- \( E_a \) represents the activation energy, the minimum energy needed for reactants to form products.- \( R \) is the gas constant, approximately 8.314 J/mol K.- \( T \) is the temperature in Kelvin.Understanding the Arrhenius equation is crucial because it explains why reactions happen faster at higher temperatures. Higher temperatures provide more energy, effectively lowering the impact of the activation energy barrier, allowing reactions to proceed more briskly. By comparing the rate constants and using this equation, we can compute activation energies to see how catalysts, like enzymes, affect reaction rates. This is key to many fields including chemistry and biology, especially in understanding enzyme efficiency.

Enzyme Catalysis

Enzymes are remarkable biological catalysts that significantly increase the rates of chemical reactions in biological systems. The enzyme carbonic anhydrase, for example, accelerates the conversion of carbon dioxide and water into bicarbonate and protons.Without the enzyme, reactions are slower due to higher activation energy barriers. When enzymes are present, they:- Lower the activation energy (\( E_a \)) needed for a reaction.- Provide an alternative reaction pathway with lower energy requirements.- Increase reaction rates immensely within living organisms.Enzymes work by binding substrates in their active sites, inducing changes that stabilize transition states. This functionality is vital in maintaining life, aiding processes like digestion, metabolism, and DNA synthesis. By reducing activation energy, enzymes ensure that necessary reactions occur swiftly and efficiently at relative body temperatures. This allows biological systems to sustain complex processes in real-time, which would otherwise take years or not occur at all.

Reaction Rate Constants

Reaction rate constants (\( k \)) serve as a numerical measure to indicate how fast a chemical reaction proceeds. A higher \( k \) value implies a faster reaction. This constant depends on temperature, as described by the Arrhenius equation, and also depends on whether catalysts are involved.For uncatalyzed reactions, rate constants tend to be lower because:- The activation energy is usually high, limiting how effectively molecules can collide and react.With enzyme catalysis, the rate constant increases substantially. In our example, the presence of carbonic anhydrase boosts the rate constant from 0.039 s\(^{-1}\) to 1.0 x 10\(^{6}\) s\(^{-1}\). This dramatic increase is due to:- The enzyme lowering the activation energy barrier.- The higher probability of successful collisions due to the enzyme's facilitation.Measuring reaction rate constants helps scientists to predict reaction behavior and to modify conditions for the desired speed of reactions. This principle not only advances our comprehension of chemical kinetics but also enhances the design of pharmaceuticals and other practical applications.