Question

The Kb for methylamine, CH3NH2, at 25 C is 4.4 x 10-4
A. Write the chemical equation for the equilibrium that corresponds to Kb.
B. By using the value of Kb, calculate ΔGo for the equilibrium in part A.
C. What is the value of ΔG at equilibrium?
D. What is the value of ΔG when [H+] = 1.6 ×10-8 M, [CH3NH3+] = 5.5 ×10-4 M, and
[CH3NH2] = 0.130 M?

Answer 1

Answer:

For A: The equation is written below.

For B: The value of standard Gibbs free energy of the reaction is 19148.5 J

For C: The value of Gibbs free energy change at equilibrium is equal to 0.

For D: The Gibbs free energy change of the reaction is -29.8 kJ

Explanation:

• For A:

The chemical equation for the dissociation of methylamine in water follows:

$CH_3NH_2(aq.)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq.)+OH^-(aq.)$

The expression for the base dissociation constant follows:

$K_b=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$

Water does not appear in the expression because it is a pure liquid.

• For B:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_b$

where,

$\Delta G^o$ = standard Gibbs free energy

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_b$ = base dissociation constant = $4.4\times 10^{-4}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (4.4\times 10^{-4})\\\\\Delta G^o=19148.5J$

Hence, the value of standard Gibbs free energy of the reaction is 19148.5 J

• For C:

At equilibrium, the Gibbs free energy change of the reaction is taken as 0.

• For D:

To calculate the concentration of hydroxide ion, we take ionic product of water. The expression of ionic product of water is written as:

$K_w=[H^+]\times [OH^-]$

We are given:

$[H^+]=1.6\times 10^{-8}M\\K_w=10^{-14}$

Putting values in above equation, we get:

$10^{-14}=1.6\times 10^{-8}\times [OH^-]$

$[OH^-]=\frac{10^{-14}}{1.6\times 10^{-8}}=6.25\times 10^{-7}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln (\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]})$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 19148.5 J

R = Gas constant = $8.314J/K mol$

T = Temperature = 298 K

$[CH_3NH_2]=0.130M$

$[CH_3NH_3^+]=5.5\times 10^{-4}M$

$[OH^-]=6.25\times 10^{-7}M$

Putting values in above equation, we get:

$\Delta G=19148.5+(8.314)(298)\ln (\frac{5.5\times 10^{-4}\times 6.25\times 10^{-7}}{0.130})\\\\\Delta G=-29789.7J$

Converting this into kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, $-29789.7J=-29.8kJ$

Hence, the Gibbs free energy change of the reaction is -29.8 kJ

Answer 2

The value of Gibbs free energy at equilibrium has been -29.8 kJ.

The $\rm K_b$ has been used for the determination of the strength of base in an ionization reaction.

The reaction can be given as:

$\rm CH_3NH_2\;(aq)\;+\;H_2O\;(l)\;\rightarrow\;CH_3NH_3^+\;(aq.)\;+\;OH^-\;(aq)$

The equation for $\rm K_b$ has been:

$\rm K_b$ = $\rm \dfrac{[CH_3NH_3^+]\;+\;[OH^-]}{[CH_3NH_2]}$

$\Delta$G standard = -RT In$\rm K_b$

$\Delta$G standard = - 8.314 $\times$ 298 K(25 $\rm ^\circ C$) $\rm \times\;In(4.4\;\times\;10^-^4)$

$\Delta$G standard = 19148.5 J

At equilibrium, the value of Gibbs free energy has been zero.

The value of $\Delta$G for the given values has been:

The concentration of hydroxide ion has to be calculated:

$\rm [OH^-]\;=\;\dfrac{k_w}{[H^+]}$

$\rm k_w$ = $\rm 10^-^1^4$

$\rm [H^+]$ = 1.6 $\rm \times\;10^-^8$ M

$\rm [OH^-]$ = $\rm \dfrac{10^-^1^4}{1.6\;\times\;10^-^8}$ M

$\rm [OH^-]$ = 6.25 $\rm \times\;10^-^7$ M

The $\Delta$G at equilibrium can be given as:

$\Delta$G = standard $\Delta$G + RT In$\rm \dfrac{[CH_3NH_3^+]\;+\;[OH^-]}{[CH_3NH_2]}$

$\Delta$G =  19148.5 J + 8.314 $\times$ 298 $\times$ $\rm In\dfrac{5.5\;\times\;10^-^4\;\times\;6.25\;\times\;10^-^7}{0.130}$

$\Delta$G = -29789.7 J

$\Delta$G = -29.8 kJ

The value of Gibbs free energy at equilibrium has been -29.8 kJ.

For more information about the Gibbs free energy, refer to the link:

https://brainly.com/question/24572190