Question

Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of $C_{2}H_{6}$, when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ?

Answer 1

To determine: The concentration of $C_{2}H_{6}$ at the equilibrium stage when being placed in a flask at  a pressure of 4.0 atm.

Given Data: The equilibrium reaction is

${ C }_{ 2 }{ H }_{ 6 }(g)\rightleftharpoons { C }_{ 2 }{ H }_{ 4 }(g)+H_{ 2 }(g)$

Initial pressure of ${ C }_{ 2 }{ H }_{ 6 }$ is 4.0 atm

Kp, the equilibrium constant, depends on partial pressures exerted by the gaseous components in the given reaction.

Kp = 0.04 atm

Formulas to be used:

Law of chemical equilibrium:

Kp is equal to the ratio of the product of  partial pressures of the products of the given reaction, each being raised to the power of their corresponding coefficients to the product of the partial pressures of the reactants, each being raised to the power of their corresponding coefficients.

Calculation:

Step 1:

Diagnose the values to be assigned in the law of chemical equilibrium

Initial concentration of ${ C }_{ 2 }{ H }_{ 6}$ = 4.0 atm

Initial concentration of ${ C }_{ 2 }{ H }_{  4}$ = 0 atm

Initial concentration of ${H }_{ 2}$ = 0 atm

Let the pressure of ${ C }_{ 2 }{ H }_{ 4}$ at equilibrium = p atm

So, the pressure of ${H }_{ 2}$ at equilibrium = p atm

The pressure of${ C }_{ 2 }{ H }_{ 6}$ at equilibrium = 4.0-p atm

Step 2:

Substitute the values obtained in the law of chemical equilibrium

${ K }_{ p }=\frac { { p }_{ { C }_{ 2 }{ H }_{ 4 } }\times { p }_{ { H }_{ 2 } } }{ { p }_{ { C }_{ 2 }{ H }_{ 6 } } }$

${ K }_{ p }=\frac { p\times p }{ 4-p } =0.04\\ { p }^{ 2 }-0.04p+0.16=0$

Step 3: Using the formula for quadratic equations, we get

$p=\frac { -0.04\pm \sqrt { { (0.04) }^{ 2 }-4\times 1\times (-0.16) }  }{ 2\times 1 } =\frac { -0.04\pm 0.80 }{ 2 }$

By taking the positive value alone,  

We get

p =0.76/2 =0.38

Hence at equilibrium, the pressure of ${ C }_{ 2 }{ H }_{ 6 }$ = 4-p-4-0.38 = 3.62 atm