Question

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

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

Answer 1

Answer: (A)

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