Question

A vessel at \[1000K\] contains $C{O_2}$ with a pressure of \[0.5{\text{ }}atm\] . Some of the $C{O_2}$ is converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is \[0.8{\text{ }}atm\] . The value of K is _______?
A) \[1.8{\text{ }}atm\]
B) \[3{\text{ }}atm\]
C) \[0.3{\text{ }}atm\]
D) \[0.18{\text{ }}atm\]

Answer 1

Hint: Using concept of dissociation, find out equilibrium partial pressure of all gases.
Then we know that, equilibrium constant (K) in terms of partial pressure of reactants and products can be written as:
\[K = \dfrac{{{P^a}_{product}}}{{{P^b}_{reactant}}}\]
Where a and b are coefficients of product and reactants respectively, while ${P_{product}}$ is partial pressure of gaseous product and ${\operatorname{P} _{reactant}}$ is partial pressure of gaseous reactant.

Complete step by step answer:
We have to write the reaction first:
\[C{O_2} + C \to 2CO\]
Here Carbon dioxide and Carbon monoxide are gases, while Carbon is graphite which is solid.
Initial pressure of $C{O_2}$ is 0.5 atm, as given in question.
This reaction takes place at constant volume and temperature of 1000K , which means that decrease in partial pressure is proportional to decrease in moles.
Now, we know that some of the carbon dioxide is converted to carbon monoxide.
This means that the number of moles of carbon dioxide present in the vessel decreases by a value, we consider to be x. Now we can say that the partial pressure of carbon dioxide will also decrease by x.
Therefore, at equilibrium, the partial pressure of $C{O_2}$ will be \[ = \left( {0.5-x} \right){\text{ }}atm\]
From reaction we know that 1 mole of $C{O_2}$ is producing 2 moles of CO
So at equilibrium, partial pressure of CO will be = 2x
Now total pressure will be addition of partial pressures of $C{O_2}$ and \[CO\]
So total pressure at equilibrium \[ = \left( {0.5-x + 2x} \right)atm\]
Also total pressure at equilibrium \[ = 0.8{\text{ }}atm\] given
Thus, both these are equal,
\[0.5 - x + 2x = 0.8 \\
0.5 + x = 0.8 \\
x = 0.8 - 0.5 \\
x = 0.3 \\\]
Now substitute the value of x, to get partial pressure of Carbon monoxide, and carbon dioxide.
Thus partial pressure of Carbon monoxide,
\[{P_{CO}} = 2x \\
= 2 \times 0.3 \\
= 0.6atm \\\]
And partial pressure of Carbon dioxide is:
\[{P_{C{O_2}}} = 0.5 - x \\
= 0.5 - 0.3 \\
= 0.2atm \\\]
We know to write equilibrium constant for the given equation:
\[K = \dfrac{{{P^a}_{product}}}{{{P^b}_{reactant}}}\]
\[C{O_2} + C \to 2CO\]
\[K = \dfrac{{{P^2}_{CO}}}{{{P_{C{O_2}}}}}\]
Now substitute the partial pressures of carbon monoxide and carbon dioxide in above equation:
\[= \dfrac{{{{\left( {0.6} \right)}^2}}}{{0.2}} \\
= \dfrac{{0.36}}{{0.2}} \\
= 1.8atm \\\]
So, thus the value of K is (A) \[1.8{\text{ }}atm\].

Note: No role of solids in calculation of K, means we should not consider solid in calculation of K, equilibrium constant.
We should take partial pressure of gases at equilibrium and not at any other point.
While taking dissociation, take care of coefficients of reactants and products.