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At equilibrium, in a test tube, the concentration of GDP is \(1 \mathrm{M}\), of GTP is \(20 \mu \mathrm{M}\), and of \(P_{i}\) is \(1 \mathrm{M}\). What is the equilibrium constant of the reaction, GTP \(\rightleftharpoons \mathrm{GDP}+\mathrm{Pi}^{\text {? }}\)

Short Answer

Expert verified
The equilibrium constant \( K_{eq} \) is \( 5 \times 10^{4} \).

Step by step solution

01

Write the Reaction

The reaction given is: \( \text{GTP} \rightleftharpoons \text{GDP} + \text{Pi} \). This is a reversible reaction, and we need to find the equilibrium constant \( K_{eq} \).
02

Express the Equilibrium Constant

The equilibrium constant for the reaction \( \text{GTP} \rightleftharpoons \text{GDP} + \text{Pi} \) is expressed using the concentrations of the products and reactants: \[ K_{eq} = \frac{[\text{GDP}][\text{Pi}]}{[\text{GTP}]} \]
03

Substitute Concentrations into the Equilibrium Expression

Substitute the given concentrations into the equilibrium constant expression: \[ K_{eq} = \frac{(1 \, \text{M})(1 \, \text{M})}{20 \times 10^{-6} \, \text{M}} \]
04

Calculate the Equilibrium Constant

Perform the calculation:\[ K_{eq} = \frac{1 \times 1}{20 \times 10^{-6}} = \frac{1}{20 \times 10^{-6}} \]This simplifies to: \[ K_{eq} = \frac{1}{20} \times 10^6 = 0.05 \times 10^6 = 5 \times 10^4 \]
05

Interpret the Result

The equilibrium constant \( K_{eq} = 5 \times 10^{4} \) means that at equilibrium, the reaction favors the formation of GDP and Pi significantly more than the remaining GTP, as indicated by a large value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reversible Reactions
Reversible reactions are chemical processes where the reactants convert to products and vice versa. These reactions are represented by a double arrow (\(\rightleftharpoons\)), indicating that the reaction can proceed in both directions. A common example is the conversion between GTP and GDP, which showcases how reversible reactions work in biological systems.

These reactions do not go to completion. Instead, they reach a state where both the forward reaction (forming products) and reverse reaction (forming reactants) occur at the same rate. This concept is essential for understanding how different substances interact and exist simultaneously in a chemical system.

The idea is not just about avoiding the complete consumption of reactants. It helps us understand that systems in chemistry attempt to reach a balance. This balance takes the form of **chemical equilibrium**, a state where molecular transactions are happening constantly, yet their concentrations remain unchanged.
Chemical Equilibrium
Chemical equilibrium is the state in a reversible reaction where the concentrations of reactants and products remain constant over time. This occurs because the rate of the forward reaction equals the rate of the reverse reaction. In the example of GTP converting to GDP and Pi, reaching equilibrium means that these substances are interchanging at an equal pace.

Equilibrium does not imply that the concentrations are equal, just that they don't change over time. Even if there is a higher concentration of some substances, the ratios remain constant. This ratio is crucial for calculating the equilibrium constant, \( K_{eq} \).

The equilibrium constant provides insight into the reactant and product balance at equilibrium. A large \( K_{eq} \) value suggests that at equilibrium, more products are formed. Conversely, a smaller \( K_{eq} \) implies that the reactants are favored. Understanding these dynamics helps chemists predict the behavior of reactions under various conditions.
Concentration Calculations
Concentration calculations are integral to determining the equilibrium constant of a reaction. They involve calculating the proportions of substances present in a system at equilibrium. For the GTP conversion example, knowing the concentration of GTP, GDP, and Pi allows us to determine the equilibrium constant \( K_{eq} \).

To find \( K_{eq} \), use the formula:
  • \[ K_{eq} = \frac{[GDP][Pi]}{[GTP]} \]
Substituting the given values into this formula is key. For example, using \([GDP] = 1 M, [Pi] = 1 M\), and \([GTP] = 20 \mu M\) changes the problem into a mathematical one.

By simplifying, you find \( K_{eq} = 5 \times 10^4 \), indicating a significant inclination toward product formation. A correct calculation requires careful attention to units (such as making sure \(\mu M\) is converted to \(M\)) and numerical operations. Mastering concentration calculations is a necessary skill for interpreting chemical equilibrium and reaction tendencies.

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Most popular questions from this chapter

The \(\mathrm{pH}\) of a \(0.15 \mathrm{M}\) propionic acid/ \(0.1 \mathrm{M}\) sodium propionate buffer is \(4.71\). What is the \(\mathrm{p} K_{\mathrm{a}}\) of propionic acid?

For a protein with a surface-exposed aspartic acid, at what \(\mathrm{pH}\) will this residue be neutral in \(75 \%\) of the protein molecules? (Assume that the \(\mathrm{p} K_{\mathrm{a}}\) is \(4.0\).)

The reaction, \(\mathrm{A}+2 \mathrm{~B} \rightleftharpoons \mathrm{C}\), has an equilibrium constant of 2000 . During a reaction, the concentration of \(\mathrm{A}\) is \(0.01\) \(\mathrm{M}\), of \(\mathrm{B}\) is \(0.2 \mathrm{M}\), and of \(\mathrm{C}\) is \(0.5 \mathrm{M}\). a. What is the reaction quotient \((Q)\) ? b. In what direction will the reaction proceed?

The \(\mathrm{p} K_{\mathrm{a}}\) of a weak acid is 5 . What is the \(\mathrm{pH}\) when the concentration of the acid form is \(0.5 \mathrm{M}\) and the concentration of the conjugate base form is \(0.05 \mathrm{M}\) ?

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