Question

For the reaction at 495 K;: \[{\Delta _r}{G^o} = - 9.478kJmo{l^{ - 1}}\]. If we start the reaction in a closed container at 495K with 22 millimoles of A, the amount of B in the equilibrium mixture is ______________________millimoles. (Round off to the nearest integer).
[R=\[8.314{\rm{ }}Jmo{l^{ - 1}}{K^{ - 1}}\];\[ln10 = 2.303\]]

• 1. 20 millimoles

2. 22 millimoles

3. 21 millimoles

4. 23 millimoles

Answer 1

Hint: Any reversible reaction will reach equilibrium in a closed system. There is no macroscopic change at this point because the overall concentrations of reactants and products remain constant.

Complete step by step solution:Equilibrium: The state in which both the reactants and products are present in concentrations that have no further tendency to change with time, resulting in no observable change in the system's properties.
The expression for Gibbs's free energy is as follows:
Where \[\Delta G = {\Delta _r}{G^o} + RT\ln K\]
\[\Delta G = \]Gibbs free energy
\[{\Delta _r}{G^o} = \]The standard change in the free energy
R= Gas Constant
T= Temperature
K= Rate constant
We know the system is in equilibrium. This implies,
\[\Delta G = 0\]and \[K \to {K_{eq}}\]
Hence, the equation associated with equilibrium will be
\[{\Delta _r}{G^o} = - RT\ln {K_{eq}}\]……..(1)
Given,
\[{\Delta _r}{G^o} = - 9.478kJmo{l^{ - 1}}\]
\[T = 495K\]
\[R = 8.314jmo{l^{ - 1}}{K^{ - 1}}\]
Putting values in equation (1), we get
\[ - 9.478kJmo{l^{ - 1}} = - 8.314Jmo{l^{ - 1}}{K^{ - 1}} \times 495K \times \ln {K_{eq}}\]
The standard change in the free energy i.e.,\[{\Delta _r}{G^o}\] value is in Kilo Joule whereas the gas constant i.e., R value is in Joules. This implies we need to convert Kilo joules to Joules.
\[{10^3} \times - 9.478kJmo{l^{ - 1}} = - 8.314Jmo{l^{ - 1}}{K^{ - 1}} \times 495K \times \ln {K_{eq}}\]
This gives, \[{K_{eq}} = 10\]
Given that there are 22 millimoles of A\[\]
Let the volume be \[V\]and x millimoles of B

\[{K_{eq}} = \dfrac{{\left[ B \right]}}{{\left[ A \right]}}\]
\[{K_{eq}} = \dfrac{{\left[ {\dfrac{x}{V}} \right]}}{{\left[ {\dfrac{{22 - x}}{V}} \right]}}\]
\[{K_{eq}} = \dfrac{{\left[ x \right]}}{{\left[ {22 - x} \right]}}\]
And we know \[{K_{eq}} = 10\]
\[10 = \dfrac{{\left[ x \right]}}{{\left[ {22 - x} \right]}}\]
\[x = 20\]
The amount of B in the equilibrium mixture is 20 millimoles.

Therefore, option (1) is right

Additional Information:Gibbs's free energy is a quantity used to calculate the maximum amount of work done in a thermodynamic system with constant temperature and pressure.

Note: A closed system cannot exchange matter, it can only exchange energy with its surroundings. We could simulate a closed system by putting a very tightly fitting lid on the pot from the previous example. A system that is isolated is one that cannot exchange matter or energy with its surroundings.