Question

Equal moles of $N{O}_2(g)$ and $NO(g)$ are to be placed in a container to produce $N_{2}O$ according to the reaction,
$N{O}_2+NO\rightleftharpoons N_{2}O+O_2;K_C=0.914$ How many moles of $N{O}_2$ and $NO$ be placed in the $5.0L$ container to have an equilibrium concentration of $N_{2}O$ to be $0.05M$ ?

• A. 0.5115
• B. 0.1023
• C. 0.0526
• D. 0.2046

Answer 1

Answer: (A)

Equilibrium concentration of $N_{2}O=O_2=0.05M$
Let the equilibrium concentration of
$N{O}_2$ and $NO$ be $xM$
$K_c=0.917=\frac{(0.05{)}^2}{x^2}$
$\Rightarrow x=0.0522M$
Initial concentrations of $N{O}_2$ and $NO$ were $0.05+x=0.1022M$ each.
Hence, moles of each gas $N{O}_2$ and $NO$ taken initially
$=5\times 0.1022=0.511$