Question

The relative equilibrium concentrations of two chemical species (a and

Answer 1

The change in standard gibbs free energy of the given reaction is   $\boxed{14.1323{\text{ kJ/mol}}}$

Further explanation:

Chemical equilibrium is the state in which concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{A(g)}} + {\text{B(g)}}\rightleftharpoons{\text{C(g)}} + {\text{D(g)}}$

Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:

${\text{K}}=\frac{{\left[{\text{D}}\right]\left[{\text{C}}\right]}}{{\left[ {\text{A}}\right]\left[{\text{B}}\right]}}$                    …… (1)

Here, K is the equilibrium constant.

The change in standard gibbs free energy $\left({\Delta{G^0}}\right)$ for the reaction is given as,

$\Delta{G^0} = -RT\ln K$                       …… (2)

Here, R is a gas constant and has the value $8.314{\text{ J/K}}\cdot{\text{mol}}$ , T is a temperature in Kelvin, and K is a equilibrium constant for the reaction.

Consider the reaction in which reactant A gives product B such as,

${\text{A}}\rightleftharpoons{\text{B}}$

It equilibrium constant can be written as follows:

$K=\frac{{\left[{\text{B}}\right]}}{{\left[{\text{A}}\right]}}$                  …… (3)

Since the ratio of [A]:[B] is 240:1, therefore, the ratio of [B]:[A] is 1:240.

Substitute this ratio in equation (3).

$\begin{aligned}K&=\frac{{\left[{\text{B}}\right]}}{{\left[{\text{A}}\right]}}\\&=\frac{1}{{240}}\\\end{aligned}$

The value of temperature is $37{\text{ }}^\circ{\text{C}}$ , thus the temperature in kelvin can be calculated as follows:

$\begin{aligned}T\left({\text{K}}\right)&=T\left({^\circ{\text{C}}}\right)+273.15\\&=37\;^\circ {\text{C}}+273.15\\&=310.15\;{\text{K}}\\\end{aligned}$

Subsititute 1/240 for equilibrium constant (K), $8.314{\text{ J/K}}\cdot{\text{mol}}$  for R and 310.15 K for T in the equation (2).

$\begin{aligned}\Delta {G^0}&= -RT\ln K\\&= -\left( {8.314{\text{ J/K}} \cdot{\text{mol}}}\right)\left({310.15{\text{ K}}}\right){\text{ln}}\left({\frac{1}{{240}}}\right)\\&=14132.3{\text{ J/mol}}\left({\frac{{{\text{1}}\;{\text{kJ/mol}}}}{{{\text{1000}}\;{\text{J/mol}}}}}\right)\\&=14.1323{\text{ kJ/mol}}\\\end{aligned}$

Therefore, change in standard gibbs free energy of the given reaction is

Learn more:

1. The difference between heat and temperature.: brainly.com/question/914750

2. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: Relative equilibrium concentrations, two chemical species, [a]:[b] = 240:1, 310.15 K, 14132.3 J/mol and 14.1323 kj/mol.

Answer 2

ΔG° = 14.1 kJ/mol

For the reaction A → B, K = [B]/[A].

If [A] = 240 and [B] = 1, then

K = 1/240 = 4.167 x 10^(-3)

The relationship between ΔG° and K is:

ΔG° = -RTlnK

where

R = the gas constant = 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

In this problem, T = (37 + 273.15) K = 310.15 K

∴ #ΔG° = -8.314 J·K^(-1)mol^(-1) × 310.15 K × ln(4.167× 10^(-3)

= -2579 × [-5.481 J·mol^(-1)] = 14 100 J·mol^(-1) = 14.1 kJ/mol

Note: We should expect ΔG° to be positive because K < 1.