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Chapter 1: Q12P (page 21)

When the reaction $A + B 鈬� C$ is at equilibrium, the concentrations of reactants are as follows: $[A] = 2 mM, [B] = 3 mM, [C] = 9 mM$ . What is the standard free energy change for the reaction?

Short Answer

Expert verified

The standard free energy change is ${螖骋}^{0} = - 18122.29 J / mol$ .

Step by step solution

01

Given data

The given reaction is,

$A + B 鈬� C$

${螖骋}^{0} = - 18122.29 J / mol$

A and Bare the substrates and C is the product.

Then,

$A + B 鈬� C$ $\begin{array}{l} [A] = 0.002 M \\ [B] = 0.003 M \\ [C] = 0.009 M \end{array}$ $\begin{array}{l} [A] = 0.002 M \\ [B] = 0.003 M \\ [C] = 0.009 M \end{array}$

02

Calculating the equilibrium constant

$\begin{array}{l} K_{e q}^{'} = \frac{[C]}{[A] 脳 [B]} \\ = \frac{[0.009 M]}{[0.002 M] 脳 [0.003 M]} \\ = 1500 M \end{array}$

03

Calculating the standard free energy change

$螖 G = 螖 G^{0} + R T \ln Q$

Where,

$螖骋$ Change in free energy

$螖 G^{0}$ Standard free energy change

R gas constant

T Standard temperature

04

At equilibrium

$\begin{array}{l} 螖 G^{0} = - R T \ln K_{e q}^{'} \\ = - 2.303 脳 8.314 J K^{- 1} m o l^{- 1} 脳 298 K 脳 \ln (1500 M) \\ = - 18122.29 J / m o l \end{array}$

Therefore, the standard free energy change is ${螖骋}^{0} = - 18122.29 J / mol$ .

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Most popular questions from this chapter

Summarize the major stages of chemical evolution.

Why is the cell membrane not an absolute barrier between the cytoplasm and the external environment?

螖骋掳鈥� for the isomerization reaction

glucose-1-phosphate (G1P) 鈫抔lucose-6-phosphate (G6P)

is 鈭�7.1 kJ 路 mol^鈭�1. Calculate the equilibrium ratio of [G1P] to [G6P] at 25掳C.

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Label the following statements true or false:

(a) A reaction is said to be spontaneous when it can proceed in either the forward or reverse direction.

(b) A spontaneous process always happens very quickly.

(c) A nonspontaneous reaction will proceed spontaneously in the reverse direction.

(d) A spontaneous process can occur with a large decrease in entropy

See all solutions

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