   Question Answers

# An equilibrium mixture of $CO(g)+{{H}_{2}}O(g)\rightleftharpoons C{{O}_{2}}(g)+{{H}_{2}}(g)$ present in a vessel of one liter capacity at ${{815}^{\circ }}C$ was found by analysis to contain 0.4 mol of CO, 0.3 mol of ${{H}_{2}}O$ , 0.2 mol of $C{{O}_{2}}$ and 0.6 mol of ${{H}_{2}}$ .a)- Calculate ${{K}_{c}}$b)- If it is derived to increase the concentration of CO to 0.6 mol by adding $C{{O}_{2}}$ to the vessel, how many moles must be added into equilibrium mixture at a constant temperature in order to get this change?(a)- \begin{align} & a)-\text{ }{{K}_{c}}=1.25 \\ & b)-\text{ a = 1}\text{.5 moles} \\ \end{align}(b)- \begin{align} & a)-\text{ }{{K}_{c}}=1 \\ & b)-\text{ a = 0}\text{.5 moles} \\ \end{align}(c)- \begin{align} & a)-\text{ }{{K}_{c}}=1.5 \\ & b)-\text{ a = 0}\text{.75 moles} \\ \end{align}(d)- \begin{align} & a)-\text{ }{{K}_{c}}=1 \\ & b)-\text{ a = 0}\text{.75 moles} \\ \end{align}  Verified
130.8k+ views
Hint: The value of equilibrium constant can be calculated by dividing the concentration of the products with the concentration of the reactant. For finding the value of the concentration of each compound the calculated number of moles should be divided by the volume of the flask.

The product of the molar concentration of the products, each raised to the power to its stoichiometric coefficient divided by the product of the molar concentration of the reactant, each raised to the power to its stoichiometric coefficient at constant temperature is called the Equilibrium constant.
a)- So, in the equation:
$CO(g)+{{H}_{2}}O(g)\rightleftharpoons C{{O}_{2}}(g)+{{H}_{2}}(g)$
${{K}_{c}}=\frac{[C{{O}_{2}}][{{H}_{2}}]}{[CO][{{H}_{2}}O]}$
Now, all the values of the compound are given in moles, it has to be converted in concentration.
It can be done by dividing the number of moles with the volume of the vessel. We get,
$[C{{O}_{2}}]=\frac{0.2}{1}=0.2$
$[{{H}_{2}}]=\frac{0.6}{1}=0.6$
$[CO]=\frac{0.4}{1}=0.4$
$[{{H}_{2}}O]=\frac{0.3}{1}=0.3$
Now, putting all the concentration in the equation, we get
${{K}_{c}}=\frac{[C{{O}_{2}}][{{H}_{2}}]}{[CO][{{H}_{2}}O]}=\frac{0.2\text{ x 0}\text{.6}}{0.4\text{ x 0}\text{.3}}=1$
So, the value of the equilibrium constant ${{K}_{c}}$ is 1.
b)- It is derived to increase the concentration of CO to 0.6 moles. This is done by adding 0.2 moles of CO at equilibrium by forcing $C{{O}_{2}}$ into the equilibrium mixture.
Let us assume that a mole of $C{{O}_{2}}$ is added into the equilibrium. Due to this addition of $C{{O}_{2}}$ the reaction will start to proceed in the backward direction to reach the equilibrium, hence the new concentrations will be:
\begin{align} & C{{O}_{2}}(g)\text{ }+\text{ }{{H}_{2}}(g)\text{ }\rightleftharpoons \text{ }CO(g)\text{ }+\text{ }{{H}_{2}}O(g) \\ & (0.2+a)\text{ 0}\text{.6 0}\text{.4 0}\text{.3} \\ & \text{(0}\text{.2+a-0}\text{.2) (0}\text{.6-0}\text{.2) (0}\text{.4+0}\text{.2) (0}\text{.3+0}\text{.2)} \\ & \text{ a 0}\text{.4 0}\text{.6 0}\text{.5} \\ \end{align}
So, putting all the new concentration into the equilibrium constant equation, we get:
$\frac{1}{{{K}_{c}}}=\frac{[CO][{{H}_{2}}O]}{[C{{O}_{2}}][{{H}_{2}}]}=\frac{0.6\text{ x 0}\text{.5}}{a\text{ x 0}\text{.4}}$
$a=0.75$
So, 0.75 mole of $C{{O}_{2}}$ must be added into equilibrium mixture at a constant temperature.
Hence, the correct answer is option (d)- \begin{align} & a)-\text{ }{{K}_{c}}=1 \\ & b)-\text{ a = 0}\text{.75 moles} \\ \end{align}

Note: The equation should be balanced for finding the number of moles. For finding the equilibrium constant the concentration should be taken. If you take the number of moles the answer would be wrong. So make sure that the number of moles is converted into concentrations. Whenever there is the addition of any substance to the equilibrium. The change will occur in the opposite direction to attain the equilibrium.