Question

Equal moles of $N{{O}_{2}}(g)$ and $N{{O}_{{}}}(g)$ are to be placed in a container to produce ${{N}_{2}}O$ according to the reaction,
$N{{O}_{2}}+NO\to {{N}_{2}}O+{{O}_{2}};\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{K}_{c}}=0.914$
How many moles of $N{{O}_{2}}$ and $NO$ be placed in 5.0 L container to have an equilibrium concentration of ${{N}_{2}}O$to be 0.05 M?
A 0.511
B 0.1023
C 0.0526
D 0.2046

Answer 1

The equilibrium constant ${{K}_{c}}$ for any reaction in equilibrium, tells us the concentration ratio of reactants and products raised to their stoichiometric coefficient.
Formula used:
${{K}_{c}}=\dfrac{{{[products]}^{a}}}{{{[reac\tan ts]}^{b}}}$ , where ‘a’ and ‘b’ are stoichiometric coefficients.

Complete answer:
We have been given a reaction that produces dinitrogen oxide, ${{N}_{2}}O$. This reaction happens in a container with equal moles of $N{{O}_{2}}(g)$ and $N{{O}_{{}}}(g)$, now we have to find the moles of $N{{O}_{2}}(g)$ and $N{{O}_{{}}}(g)$ used in a 5.0 L container to reach the concentration of ${{N}_{2}}O$to 0.05 M.
For this we have to see the reaction, $N{{O}_{2}}+NO\to {{N}_{2}}O+{{O}_{2}}$
We will assume that initially, when no reaction has happened, the concentration of reactants is ‘x’, and products is ‘0’ , and when equilibrium has reached, the concentration of reactants will be $\dfrac{x-y}{5}$ and that of products will be $\dfrac{y}{5}$ . So,

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{{O}_{2}}+NO\to {{N}_{2}}O+{{O}_{2}} \\\ & initially\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \\\ & equilibrium\,\,\,\dfrac{x-y}{5}\,\,\,\dfrac{x-y}{5}\,\,\,\,\,\,\dfrac{y}{5}\,\,\,\,\,\,\,\,\,\dfrac{y}{5} \\\ \end{aligned}$$ Now, we are given that concentration of ${{N}_{2}}O$ should be 0.05, so concentration of ${{N}_{2}}O$in the product will be, $[{{N}_{2}}O]=\dfrac{y}{5}=0.05\,M$, so we will get$y=0.25\,moles$. Now keeping these quantities in the formula of equilibrium constant, $${{K}_{c}}=\dfrac{[{{N}_{2}}O][{{O}_{2}}]}{[N{{O}_{2}}][NO]}$$ given that, $${{K}_{c}}=0.914$$ 0.914 = $\dfrac{0.25\times 0.25}{{{(x-0.25)}^{2}}}$ So, we will get x = 0.511 moles. Hence, option A is correct, as 0.511 moles of $N{{O}_{2}}(g)$ and $N{{O}_{{}}}(g)$will be needed. **Note:** We have got the total concentration of ${{N}_{2}}O$ to be 0.05 M, so, this will be divided by the volume to get the number of moles of ${{N}_{2}}O$. So, through number of moles of ${{N}_{2}}O$, we can calculate the remaining number of moles of reactants, $N{{O}_{2}}(g)$ and $N{{O}_{{}}}(g)$which is denoted as x.