Question

In which of the following reaction product is more stable :-
(1) N, + 3H, = 2NH2 ; K, = 2.3 x 10-2 (2) N, + 0,= 2N0 ; K, = 2 x 102
song = 2 - Ci+? =2-
(3) H, + 12 = 2HI ; K, = 294
(4) XeO + 30+ F, = XeO,F2 ; Kq=1.4 x 10-3
​

Answer 1

Answer:

XeF6 +H,0 =

XeOFq + 2HF constant = K,, XeO4 + XeF

XeOFq + XeO, F, constantă

XeO,F2 + H2O will be

K2. Then equilibrium constant for the reaction XeO, + 2HF

Ki

K₂

K,

(4) (K₂²

(3) K

(1) K2

(2) K, + K​

Answer 2

The correct answer is option (3). $H_{2} + I_{2}$ ⇄ $2HI$, $K_3 = 294.$

Given:

The four reversible reactions are given.

To Find:

We have to find out which of the given reaction products is more stable.

Solution:

Consider a reversible chemical reaction, $A + B$ ⇄ $C + D.$

Then, the rate constant or  rate coefficient, K, of the reaction is given by,

$K = \frac{Concentration of products}{Concentration of reactants} = \frac{[C].[D]}{[A]. [B]}$.

The products of a reaction are more stable than the reactants in a reaction having a large value of K.

Consider (1). $N_{2} + 3H_{2}$ ⇄ $2NH_{3}$.

Here, the value of K = $2.13$ × $10^{-2} = 0.023.$

Consider (2). $N_{2} + O_{2}$ ⇄ $2NO$.

The value of K = $2$ × $10^{2} = 200.$

Consider (3). $H_{2} + I_{2}$ ⇄ $2HI$.

The value of K = $294.$

Consider (4). $XeO + \frac{1}{2} O_{2} + F_{2}$ ⇄ $XeO_{2}F_{2}$.

Here, the value of K = $1.4$ × $10^{-3} = 0.0014.$

Among these, the value of K is higher for option (3).

Hence, among the given reaction products, the more stable one is (3).  $H_{2} + I_{2}$ ⇄ $2HI$.

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