
If $A{B_2}$ dissociates as $A{B_2} \Leftrightarrow AB(g) + B(g)$. When the initial pressure of$A{B_2}$is 600mm of Hg, the total equilibrium pressure is 800 mm of Hg. Calculate ${K_p}$ for the reaction, assuming that the volume of the system remains unchanged.
A. 50
B. 100
C. 166.8
D. 400
Answer
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Hint: When equilibrium constant is expressed in terms of the partial pressure of the reactants and products then it is denoted by ${K_P}$. Mathematically it is written as:
${K_P} = \dfrac{{P_X^x \times P_Y^y}}{{P_A^a \times P_B^b}}$
Here $P_X^x, P_Y^y, P_A^a{\text{ }} and {\text{ }}P_B^b$ are the partial pressure of products and reactants respectively.
Complete step by step answer:
As per the question, the initial pressure of $A{B_2}$ is given as 600 mm of Hg and the total equilibrium pressure is 800 mm of Hg.
The equilibrium reaction is given as:
$A{B_2} \Leftrightarrow AB(g) + B(g)$
$600 \;\; \;\; \; \;\; \;\; \;\; 0 \;\;\;\;\;\;\;\; \;\; \;\; 0 \;\; \;\;\; Initial$
$600-x \;\; \;\; x \;\; \;\; \;\;\;\; \;\; \;\; \;\; x \;\; \;\;\;At \;equilibrium$
Where x is the pressure of $AB$ and $B$ at equilibrium.
The total pressure at equilibrium is equal to the sum of the partial pressure of $A{B_2}$, $AB$ and $B$. Thus the total pressure at equilibrium = $600 - x + x + x$ ... (1)
Given that the total pressure at equilibrium = 800 mm of Hg. (2)
Thus the equation (1) is equal to equation (2). Hence
$600 - x + x + x = 800$
$ \Rightarrow 600 + x = 800$
$ \Rightarrow x = 800 - 600 = 200$mm of Hg.
Thus the partial pressure $A{B_2}$ at equilibrium is calculated as
$600 - x = 600 - 200 = 400$mm of Hg.
Similarly, the partial pressure of $AB$ and $B$ at equilibrium is 200 mm of Hg respectively. The equilibrium constant ${K_P}$ for the above case be represented as:
${K_P} = \dfrac{{{P_{AB}} \times {P_B}}}{{{P_{A{B_2}}}}}$
On putting the value partial pressure of $A{B_2}$, $AB$ and $B$ at equilibrium in the expression of ${K_P}$, we get:
${K_P} = \dfrac{{200 \times 200}}{{400}}$
$ \Rightarrow {K_P} = \dfrac{{40000}}{{400}} = 100$
So, the correct answer is Option B .
Note: Equilibrium constant in terms of partial pressure, i.e. ${K_P}$ is a dimensionless quantity because nowadays activities instead of the pressures is used which is pressure represented with respect to standard state pressure and they cancel out the unit of each. Similarly, the equilibrium constant in terms of concentration is also a dimensionless quantity however the magnitude of the equilibrium constant either in terms of pressure or in terms of concentration both depends upon the standard state chosen.
${K_P} = \dfrac{{P_X^x \times P_Y^y}}{{P_A^a \times P_B^b}}$
Here $P_X^x, P_Y^y, P_A^a{\text{ }} and {\text{ }}P_B^b$ are the partial pressure of products and reactants respectively.
Complete step by step answer:
As per the question, the initial pressure of $A{B_2}$ is given as 600 mm of Hg and the total equilibrium pressure is 800 mm of Hg.
The equilibrium reaction is given as:
$A{B_2} \Leftrightarrow AB(g) + B(g)$
$600 \;\; \;\; \; \;\; \;\; \;\; 0 \;\;\;\;\;\;\;\; \;\; \;\; 0 \;\; \;\;\; Initial$
$600-x \;\; \;\; x \;\; \;\; \;\;\;\; \;\; \;\; \;\; x \;\; \;\;\;At \;equilibrium$
Where x is the pressure of $AB$ and $B$ at equilibrium.
The total pressure at equilibrium is equal to the sum of the partial pressure of $A{B_2}$, $AB$ and $B$. Thus the total pressure at equilibrium = $600 - x + x + x$ ... (1)
Given that the total pressure at equilibrium = 800 mm of Hg. (2)
Thus the equation (1) is equal to equation (2). Hence
$600 - x + x + x = 800$
$ \Rightarrow 600 + x = 800$
$ \Rightarrow x = 800 - 600 = 200$mm of Hg.
Thus the partial pressure $A{B_2}$ at equilibrium is calculated as
$600 - x = 600 - 200 = 400$mm of Hg.
Similarly, the partial pressure of $AB$ and $B$ at equilibrium is 200 mm of Hg respectively. The equilibrium constant ${K_P}$ for the above case be represented as:
${K_P} = \dfrac{{{P_{AB}} \times {P_B}}}{{{P_{A{B_2}}}}}$
On putting the value partial pressure of $A{B_2}$, $AB$ and $B$ at equilibrium in the expression of ${K_P}$, we get:
${K_P} = \dfrac{{200 \times 200}}{{400}}$
$ \Rightarrow {K_P} = \dfrac{{40000}}{{400}} = 100$
So, the correct answer is Option B .
Note: Equilibrium constant in terms of partial pressure, i.e. ${K_P}$ is a dimensionless quantity because nowadays activities instead of the pressures is used which is pressure represented with respect to standard state pressure and they cancel out the unit of each. Similarly, the equilibrium constant in terms of concentration is also a dimensionless quantity however the magnitude of the equilibrium constant either in terms of pressure or in terms of concentration both depends upon the standard state chosen.
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