Question

The equilibrium constant for the isomerisation of butane at 25$^ \circ C$ is ${K_C}$ = 8.
Butane → Isobutane
If 5.8 g butane is introduced into a 12.5 L vessel at 25$^ \circ C$. What mass of isobutene will be present at equilibrium?
(A) 5.15 g
(B) 2.56 g
(C) 7.49 g
(D) 5.80 g

Answer 1

Hint: First we will calculate the moles of butane present initially. Also ${K_C}$ is the ratio of molar concentration of products to reactants at equilibrium concentrations raised to the power of their stoichiometric coefficients.
For a reaction: $aA \rightleftharpoons bB$
The value of ${K_C}$ for the above reaction will be:
${K_C} = \dfrac{{{{\left[ B \right]}^b}}}{{{{\left[ A \right]}^a}}}$
Similarly we will write the expression of ${K_C}$ for the isomerisation reaction and calculate the concentration of isobutane at equilibrium. Then from it find the weight if isobutane is produced.

Complete step by step solution:
-First of all we will calculate the initial moles of butane (${C_4}{H_{10}}$).
According to the question the given weight of butane is = 5.8 g; and we know that molecular weight of 58 g/mol.
We will now calculate the number of moles: $n = \dfrac{W}{M}$ = $\dfrac{{5.8}}{{58}}$ = 0.1
So, the number of moles of butane present initially are 0.1 moles.
-From the isomerisation reaction:
                                              Butane → Isobutane
Initial concentration: 0.1 0
Final concentration: 0.1 – x x
-We know that ${K_C}$ is basically the ratio of equilibrium concentrations of the products over the equilibrium concentrations of the reactants each raised to the power of their stoichiometric coefficients.
For the given reaction the value of ${K_C}$ is given to be = 8 and in the form of expression it is written as:
 ${K_C} = \dfrac{{\left[ P \right]}}{{\left[ R \right]}}$ = 8
 8 = $\dfrac{x}{{0.1 - x}}$
 8 ( 0.1 – x ) = x
 0.8 – 8x = x
0.8 = x + 8x
 0.8 = 9x
 x = $\dfrac{{0.8}}{9}$ = 0.0888
So, from the reaction we can say that the concentration of isobutane at equilibrium is 0.0888 moles.
-Using the moles of isobutane obtained above we will calculate the weight of isobutane produced.
The molecular weight of isobutane is 58 g/mol and the moles are 0.0888.
$n = \dfrac{W}{M}$
0.0888 = $\dfrac{W}{{58}}$
W = 0.0888 × 58 = 5.1504 g
Hence we can conclude that the amount of isobutane produced is 5.1504 g.

So, the correct option is: (A) 5.15 g

Note: Equilibrium constant for any reaction depends on temperature and is independent of the actual quantities of reactants and products, the presence of catalyst and the presence of any inert material. It also does not depend on the concentrations, pressures and volumes of the reactants and products.