A flask is initially filled with pure $N_{2} O_{g} (g)$ having pressure 2 bar and following equilibria are established.
$N_{2} O_{3} (g) ⇌ {NO}_{2} (g) + NO (g)$ $K_{P_{1}} = 2.5 bar$ ?
$2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g)$ $K_{P_{y}} =$ ?
If at equilibrium partial pressure of NO(g) was found to be 1.5 bar, then:

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

Equilibrium partial pressure of $N_{2} O_{3} (g)$ is 0.5 bar

Equilibrium partial pressure of ${NO}_{2} (g)$ is 0.83 bar

Equilibrium partial pressure of $N_{2} O_{4}$ is 0.33 bar

Value of $K_{P_{2}}$ is 0.48 bar

answer is A, B, D, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here, we have,

$N_{2} O_{3} ⟶ {NO}_{2} + NO$

$\begin{array}{ll} 2 - x x - y x \\ 2 {NO}_{2} ⟶ N_{2} O_{4} \\ x - y y / 2 \end{array}$

$x = 1.5$ (given)

$\begin{array}{l} K_{P_{1}} = \frac{(3 - y) \times 1.5}{0.5} \\ P_{N_{2} O_{5}} = 2 - 1.5 = 0.5 bar \\ P_{{NO}_{2}} = 0.83 bar \\ y = 1.5 - 0.83 = 0.67 \\ P_{N_{2} O_{4}} = 0.3 bar \\ K_{P_{2}} = \frac{0.33}{(0.82)^{2}} \\ = 0.48 \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

Get Expert Academic Guidance – Connect with a Counselor Today!