Question: The reaction is the following to calculate the number of moles \[C{{O}_{2}}+NO \to N{{O}_{2}}+CO\]...

Question

The reaction is the following to calculate the number of moles
$C{{O}_{2}}+NO \to N{{O}_{2}}+CO$
K=64. One mole of each gas is taken in 1 liter of flask. Calculate number of moles of NO and CO at equation.

Answer 1

Solution

The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of the reactants. If the value of equilibrium constant is less than 1 then the reaction will be moving towards the left. If the equilibrium constant is more than 1 then the reaction will move towards right. If the equilibrium constant is equal to 1 then the reaction is at equilibrium.

Complete answer:
If the reaction has some solid and liquid components then their concentration is considered as nil or zero. Now let us consider the concentration of all the gases in reactant and products as 1M each. The moles of the gases are ‘x’. The equilibrium concentration of the gases will be
1. $C{{O}_{2}}$ - $(1-x)$
2. $NO-(1-x)$
3. $N{{O}_{2}}-(1+x)$
4. $CO-(1+x)$
The expression we get for the equilibrium constant is
$K=\dfrac{[N{{O}_{2}}][CO]}{[C{{O}_{2}}][NO]}$ (K=64 given in the question)
Now substituting the concentrations
$64=\dfrac{(1+x)(1+x)}{(1-x)(1-x)}$
Taking under root both sides we get
$8=\dfrac{(1+x)}{(1-x)}$
On simplifying we get
$8-8x=1+x$
On taking the 1 in left hand side and taking 8x in right hand side we get,
$7=9x$
On simplifying we get,
$x=\dfrac{7}{9}$
Now the moles of NO is (1-x). So now we will substitute the value of x in it. We get,
$=1-\dfrac{7}{9}$
On further solving,
= $\dfrac{2}{9}$
Now the moles of CO is (1+x). So now we will substitute the value of x in it. We get,
$=1+\dfrac{7}{9}$
on further solving we get,
$=\dfrac{16}{9}$
So the correct answer for this question is NO is $\dfrac{2}{9}$ and for CO is $\dfrac{16}{9}$ .

Note: When the equilibrium constant is greater than 1 then it means the concentration of product is greater than reactant. If the equilibrium is less than one then the concentration of reactant is higher than products. If the equilibrium is equal to one then the reactants and products concentration is the same.