Question

Two solids dissociated as follows
${ A(s)\rightleftharpoons B(g)+C(g);\quad { K }_{ p_{ 1 } } }={ x atm }^{ 2 }$
${ D(s)\rightleftharpoons C(g)+E(g);\quad { K }_{ p_{ 2 } } }={ y atm }^{ 2 }$
The total pressure when both the solids dissociate simultaneously is:
a.) ${ x }^{ 2 }{ +y }^{ 2 }{ atm }$
b.) $\sqrt { { x }^{ 2 }-{ y }^{ 2 } } { atm }$
c.) ${ 2\sqrt { x+y } }{ atm }$
d.) ${ \sqrt { x+y } }{ atm }$

Answer 1

Hint: According to the ideal gas equation, “pressure is directly proportional to the concentration, at constant temperature and volume.” So, we can write our equilibrium pressure in terms of the partial pressure of the gas.

Complete step by step answer:
According to the ideal gas equation, we have
${ PV=nRT }$
or ${ P=n/VRT }$
or ${ P=cRT }$ [Since, { c=n/V }]
where, P = pressure
V = volume
n= number of moles
R = gas constant
T = temperature in Kelvin
c = concentration

The total pressure at equilibrium;
${ A(s)\rightleftharpoons B(g)+C(g);\quad { K }_{ p_{ 1 } } }={ x atm }^{ 2 }$
 where, ${ K }_{ p }$ =equilibrium constant
So, ${ K }_{ p }={ P }_{ B }+{ P }_{ C }$
${ D(s)\rightleftharpoons C(g)+E(g);\quad { K }_{ p_{ 2 } } }={ y atm }^{ 2 }$
${ K }_{ p }{ ={ P }_{ c } }{ +P }_{ E }$

Since these two reactions are occurring simultaneously and both A and D dissociates into C, so, C will become ${ P }_{ 1 }{ +P }_{ 2 }$
Hence, the equilibrium constant for both the reactions can be written as;

${ x={ P }_{ 1 }({ P }_{ 1 } }{ +P }_{ 2 }{ ) }$……………….(1)
${ y=({ P }_{ 1 } }{ +P_{ 2 } }){ P }_{ 2 }$……………(2)

Now, adding equation 1 and 2;
${ x+y=P }_{ 1 }{ (P }_{ 1 }{ +P }_{ 2 }{ )(P }_{ 1 }{ +P }_{ 2 }{ )P }_{ 2 }$
${ x+y =({ P }_{ 1 } }{ +P }_{ 2 }{ ) }{ +(P }_{ 1 }{ +P }_{ 2 }{ ) }$
${ x+y=({ P }_{ 1 } }{ +P }_{ 2 })^{ 2 }$
The total pressure will be given below;
${ P }_{ T }{ =P }_{ B }{ +P }_{ C }{ +P }_{ E }$
= ${ 2({ P }_{ 1 } }{ +P }_{ 2 }{ ) }$
${ P }_{ T }={ 2\sqrt { x+y } }{ atm }$
So, the correct answer is “Option C”.

Additional Information:
The partial pressure of a gas is a measure of the thermodynamic activity of the gas molecules.
Gases dissolve, diffuse and react according to their partial pressures, and not according to their concentration in gas, mixtures, or liquids.
For example: If the vessel is filled with different non-reactive gases in the state of thermal equilibrium, the total pressure of the mixture of the gases is equal to the sum of pressure of individual gases.
Dalton’s Law:- According to this law, “the total pressure of a mixture of gases is equal to the sum of the partial pressure of the individual gases in the mixture.”

Note: The possibility to make a mistake is that you may take the concentration but we have to take the partial pressures of the gases. Also, Dalton’s Law is only applicable when the gases in the mixture do not react with each other.