Problem 86 - Chemistry The Central Science In Si Units

Question

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) 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 \(21^{\circ} \mathrm{C} ?\) (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?

Step-by-Step Solution

Answer

Urease lowers the activation energy significantly compared to the uncatalyzed reaction, making the reaction faster even at lower temperatures.

Step 1: Write the Balanced Equation

The reaction of urea with water, catalyzed by the enzyme urease, can be written as: \[\text{(NH}_2\text{CONH}_2\text{)} + \text{H}_2\text{O} \rightarrow \text{CO}_2 + 2\text{NH}_3\] This equation shows that one molecule of urea reacts with one molecule of water to produce one molecule of carbon dioxide and two molecules of ammonia.

Step 2: Calculate Difference in Activation Energy

The rate constant for the uncatalyzed reaction at \(100^{\circ} \text{C}\) is \(4.15 \times 10^{-5} \text{s}^{-1}\). For the catalyzed reaction at \(21^{\circ} \text{C}\) it is \(3.4 \times 10^4 \text{s}^{-1}\). Assuming the rate of the catalyzed reaction at \(100^{\circ} \text{C}\) is the same as at \(21^{\circ} \text{C}\), the difference in activation energy \( \Delta E_a \) between the catalyzed and uncatalyzed reactions can be determined using the Arrhenius equation: \[\ln\left( \frac{k_2}{k_1} \right) = \frac{E_{a1} - E_{a2}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)\] where \( k_2 = 3.4 \times 10^4 \text{s}^{-1} \), \( k_1 = 4.15 \times 10^{-5} \text{s}^{-1} \), and \( T_1 = T_2 \) since they are assumed at the same temperature. Thus, the difference in activation energies \( \Delta E_a \) can be calculated from this equation.

Step 3: Compare Reaction Rates at Different Temperatures

When comparing the rate of the catalyzed reaction at \(100^{\circ} \text{C}\) versus \(21^{\circ} \text{C}\), the rate is expected to be much higher at \(100^{\circ} \text{C}\) because the reaction rate generally increases with temperature for enzyme-catalyzed reactions. This implies that the actual activation energy for the catalyzed reaction at \(100^{\circ} \text{C}\) is likely different from the value at \(21^{\circ} \text{C}\).

Step 4: Conclude on Activation Energy Differences

Taking into account that the enzyme-catalyzed reaction at higher temperatures leads to higher rates, it can be concluded that the activation energy of the catalyzed reaction is significantly lowered compared to the uncatalyzed reaction. This is evident because even at lower temperatures (21°C), the enzyme significantly increases the rate. This suggests the enzyme drastically reduces the activation energy required for the reaction compared to its uncatalyzed counterpart.

Key Concepts

Rate Constants in Enzyme-Catalyzed Reactions

Rate constants play a critical role in understanding how fast a chemical reaction proceeds. They provide a numerical value that represents the speed at which reactants turn into products. In the context of enzyme catalysis, rate constants can show the dramatic effect enzymes have on reaction rates.

For the reaction of urea with water in the absence of urease, the rate constant is given as a very small value: \(4.15 \times 10^{-5} \, \text{s}^{-1}\) at \(100^{\circ} \text{C}\). This slow rate reflects the difficulty in breaking down urea without the catalyst's help. However, when urease, the enzyme, is present, the rate constant jumps dramatically to \(3.4 \times 10^4 \, \text{s}^{-1}\) at a much lower temperature of \(21^{\circ} \text{C}\).

This comparison shows how enzymes can increase reaction rates significantly, exemplified by the thousand-fold increase in the rate constant. The key takeaway is that enzymes are effective in speeding up reactions under mild conditions.

Understanding Activation Energy

Activation energy is a crucial concept that helps explain why some reactions happen quickly while others proceed more slowly. It is the minimum energy required for the reactants to transform into products during a chemical reaction. Enzymes, like urease, substantially lower this activation energy, making reactions much more feasible.

In the urease-catalyzed reaction, when comparing reactions conducted at different temperatures, it's crucial to understand their activation energy differences. Even at lower temperatures, the enzyme increases the reaction rate, indicating that it has lowered the activation energy barrier that would normally slow down the process.

This reduction in activation energy is essential in biological systems, allowing complex biochemical reactions to occur rapidly and efficiently at body temperature. The enzyme provides an alternative pathway for the reaction with lower activation energy, facilitating a faster reaction.

Demystifying the Arrhenius Equation

The Arrhenius equation is a key tool used to understand the dependencies of reaction rates on temperature and activation energy. It is formulated as:

\[ k = A e^{- \frac{E_a}{RT}} \]

where:

• \(k\) is the rate constant
• \(A\) is the pre-exponential factor
• \(E_a\) is the activation energy
• \(R\) is the gas constant
• \(T\) is the temperature in Kelvin

In essence, the Arrhenius equation shows that the rate constant \(k\) increases exponentially with an increase in temperature \(T\) or a decrease in activation energy \(E_a\). This explains why enzyme reactions tend to proceed at a much higher rate even at lower temperatures, as enzymes effectively reduce \(E_a\).

By using this equation, scientists can calculate how changes in temperature or enzyme presence affect the speed of reactions. It reveals why reactions accelerate with temperature and how enzymes importantly lower the activation energy required, enhancing the reaction rate significantly.