Question

$2NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+{Br}_2\left(g\right)$. At equilibrium if nitrosyl bromide ($NOBr$) is $40$% dissociated at certain temp. and a total pressure of $0.30atm$. $K_p$ for the reaction $2NO\left(g\right)+{Br}_2\left(g\right)\rightleftharpoons 2NOBr\left(g\right)$ is:

• A. $45$
• B. $25$
• C. $0.022$
• D. $0.25$

Answer 1

The correct option is C $0.022$

The reaction;

$2NOBr\rightleftharpoons 2NO+B{r}_2$

initial (atm) $x$ $0$ $0$

change(atm) $-2y$ $+2y$ $+y$

equilibrium(atm) $x-2y$ $2y$ $y$

initial(atm) 2 0 0

change(atm) 2(0.4) 2(0.4) 0.4

equlibrium (atm) 1.2 0.8 0.4

$n_T$ =2.4

$K_P=\frac{P_{NO}^2\times P_{{Br}_2}}{(P_{NOBr}{)}^2}=\frac{(x_{NO}P{)}^2\times x_{B{r}_2}P}{(x_{NOBr}P{)}^2}$

$K_P=\frac{(\frac{0.8}{2.4}\times 0.3{)}^2\times \frac{0.4}{2.4}\times 0.3}{(\frac{1.2}{2.4}\times 0.3{)}^2}=0.022$.

Hence, 0.022 is the answer.