Question

The aqueous reaction: L-glutamate \(+\) pyruvate \(\rightleftarrows \alpha\) -ketoglutarate \(+\mathrm{L}\) -alanine is catalyzed by the enzyme \(\mathrm{L}\) -glutamate-pyruvate aminotransferase. At \(300 \mathrm{~K},\) the equilibrium constant for the reaction is 1.11 . Predict whether the forward reaction will occur if the concentrations of the reactants and products are [L-glutamate] \(=3.0 \times 10^{-5} \mathrm{M}\), [pyruvate] \(=3.3 \times 10^{-4} M,[\alpha\) -ketoglutarate \(]=1.6 \times 10^{-2} M\), and \([\mathrm{L}\) -alanine \(]=6.25 \times 10^{-3} \mathrm{M}\)

Short answer

The forward reaction is not favored as the reverse reaction is favored to reach equilibrium.

Step-by-step solution

Write the Reaction Equation

The reaction is given as: L-glutamate + pyruvate \( \rightleftarrows \alpha \)-ketoglutarate + L-alanine.

Identify Given Concentrations

Identify the concentrations given in the problem: - \([\text{L-glutamate}] = 3.0 \times 10^{-5} \, M\)- \([\text{pyruvate}] = 3.3 \times 10^{-4} \, M\)- \([\alpha\text{-ketoglutarate}] = 1.6 \times 10^{-2} \, M\)- \([\text{L-alanine}] = 6.25 \times 10^{-3} \, M\).

Calculate the Reaction Quotient, Q

The reaction quotient \( Q \) is given by the formula:\[ Q = \frac{[\alpha\text{-ketoglutarate}] \, [\text{L-alanine}]}{[\text{L-glutamate}] \, [\text{pyruvate}]} \]Substituting the values, we get:\[ Q = \frac{(1.6 \times 10^{-2}) \, (6.25 \times 10^{-3})}{(3.0 \times 10^{-5}) \, (3.3 \times 10^{-4})} \approx 1010 \]

Compare Q with Equilibrium Constant, K

The equilibrium constant for the reaction is given as \( K = 1.11 \). We compare the calculated \( Q \approx 1010 \) to \( K = 1.11 \).

Interpretation

Since \( Q \gg K \), this indicates that the reaction is producing more products than predicted by the equilibrium constant under the initial conditions. Thus, the reverse reaction is favored to re-establish equilibrium.

Key concepts

Equilibrium Constant

One of the key concepts in understanding chemical reactions is the equilibrium constant, denoted as \( K \). The equilibrium constant of a reaction provides a snapshot of the ratio of product concentrations to reactant concentrations at equilibrium, where the system is in a stable state and no net reaction occurs.
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, which means that the concentrations of the products and reactants remain constant over time. The equilibrium constant is specific to each chemical reaction and is affected by changes in temperature.
In many reactions, a larger \( K \) value indicates a tendency to favor the formation of products, while a smaller \( K \) implies the predominance of reactants at equilibrium. For the reaction in the exercise, \( K = 1.11 \) suggests that neither the reactants nor the products are overwhelmingly favored. Knowing the \( K \) helps predict the behavior of a system when not at equilibrium.

Reaction Quotient

The reaction quotient, \( Q \), works similarly to the equilibrium constant, but it applies to reaction mixtures that are not necessarily at equilibrium. \( Q \) is calculated using the same formula as \( K \), which compares the concentrations of products to reactants.
To determine the direction in which a reaction will proceed, \( Q \) is compared with \( K \):

• If \( Q < K \), the forward reaction is favored, meaning that more reactants will turn into products until equilibrium is reached.
• If \( Q > K \), as was calculated to be approximately 1010 in the exercise, the reverse reaction is favored, indicating an excess of products that will convert back to reactants.
• If \( Q = K \), the system is already at equilibrium.

In the given problem, \( Q \) being much larger than \( K \) reveals that the system currently has too much product and will shift towards the reactants to achieve equilibrium again.

Enzyme Catalysis

Enzymes play a critical role in facilitating chemical reactions, especially in biological systems. Enzyme catalysis is the process by which enzymes accelerate reactions. They do so by lowering the activation energy required for the reaction to proceed.
In the discussed reaction, the enzyme L-glutamate-pyruvate aminotransferase acts as a catalyst. While the equilibrium constant (\( K \)) itself isn’t altered by the enzyme, the rate at which equilibrium is achieved gets faster.
Through binding specific substrates in their active sites, enzymes provide an environment conducive to chemical change. This specificity ensures that the reaction path is efficiently directed towards the desired products.
Enzyme catalysis is vital in maintaining homeostasis within organisms by regulating the speed of essential biochemical reactions without which life could not sustain.

Reversible Reactions

Reversible reactions are chemical processes where products can revert to the original reactants under suitable conditions. This ability to go back and forth is indicated by the double-headed arrow (\( \rightleftarrows \)) in the reaction equation.
In a reversible reaction, both the forward and reverse reactions occur simultaneously. The relative concentrations of the reactants and products at any given time determine the predominant direction of the reaction. At equilibrium, both pathways occur at the same rate, though not necessarily with equal concentrations of reactants and products.
This is evident when examining \( Q \) in relation to \( K \), which helps us predict shifts in the reaction. In biological systems, reversible reactions allow for flexibility and fast response to changing cellular conditions, making them crucial to metabolic pathways and regulatory mechanisms.
Understanding these concepts assists in predicting how a system might behave in response to various changes in the reaction conditions, enabling a more controlled manipulation of chemical processes.