Question

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

• A. 45
• B. 25
• C. 0.022
• D. 0.25

Answer 1

Answer: (A)

45

$2NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+{Br}_2\left(g\right)$

At $t=0$ $a$ $0$ $0$

At $a(1-\alpha )$ $2a\alpha /2$ $a\alpha /2$

$t=t_{eq}$

$\alpha =0.4$ ($\because$ since $40$% dissociated)

total pressure $=0.30$ atm

$K_P=\frac{{\left(PNO\right)}^2\times \left({PBr}_2\right)}{{\left(PNOBr\right)}^2}$

Partial pressure of $NO=\frac{a\alpha }{a(1+\alpha /2)}\times P=\frac{0.4}{1.2}\times 0.3=0.1$

Partial pressure of ${BR}_2=\frac{a\alpha /2}{a(1+\alpha /2)}\times P=0.05$

Partial pressure of $NOBr=\frac{a(1-\alpha )}{a(1+\alpha /2)}\times P=0.15$

$K_{P_1}=\frac{{\left(0.1\right)}^2\times 0.05}{{\left(0.15\right)}^2}=\frac{1}{45}$

$K_{P_2}$ for $2NO+{Br}_2\rightleftharpoons 2NOBr$

$K_{P_2}={KP}_1^{-1}=45$