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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}$

Answer

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130.8k+ views

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.

& a)-\text{ }{{K}_{c}}=1 \\

& b)-\text{ a = 0}\text{.75 moles} \\

\end{align}$