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The Equilibrium Constant for the Following Reaction is 1.6 ×105 at 1024k Find the Equilibrium Pressure of All Gases If 10.0 Bar of Hbr is Introduced into a Sealed Container at 1024k - Chemistry

The equilibrium constant for the following reaction is 1.6 ×105 at 1024K

H2(g) + Br2(g) ⇌ 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K

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Solution 1

Given,

`K_P` for the reaction i.e., `H_(2(g)) + Br_(2(g)) "↔" 2HBr_(g)` is `1.6 xx 10^5`

Therefore, for the reaction `2HBr_(g) "↔" H_(2(g)) + ZBr_(2(g))`  the equilibrium constant will be,

`K_P^' = 1/K_P`

`= 1/(1.6 xx 10^5)`

`= 6.25 xx 10^(-6)`

Now, let p be the pressure of both H2 and Br2 at equilibrium.

                       `2Hbr_(g) "↔" H_(2(g)) + Br_(2(g))`

Initila conc          10        0           0

At equilibrium   10-2p     p           p

Now, we can write,

`(p_(H_2Br_2) xx p)/p_(HBr)^2 = K_P^'`

`(pxxp)/(10-2p)^2 = 6.25 xx 10^(-6)`

`p/(10-2p) = 2.5 xx 10^(-3)`

`p =2.5 xx 10^(-2) - (5.0 xx 10^(-3))p`

`(1005 xx 10^(-3))p = 2.5 xx 10^(-2)`

`p = 2.49 xx 10^(-2) bar = 2.5 xx 10^(-2)` bar (approximately)

Therefore, at equilibrium,

`[H_2] = [Br_2] = 2.49 xx 10^(-2)` bar

[HBr] = `10 - 2 xx (2.49 xx 10^(-2))` bar

= 9.95 bar = 10 bar (approximately)

Solution 2

Step I Caculation of `K_p`

`H_2 (g) + Br_2 (g) ⇌ 2HBr(g)`

`K_p = K_c(RT)^(trianglen) = K_c(RT)^0`

(`:. triangle_n = 2 - 2` = zero)

Step II Calculation of partial pressure of gases

                      `2HBr(g) ⇌  H_2(g) + Br_2(g)`

Initial Pressure  10 bar       zero         zero

Eqm pressure   (10- P) bar  P/2 bar     P/2 bar

`K_p^' = (pH_2xxpBr_2)/(p^2HBr)  or 1/(1.6 xx  10^5) = (P/2 xx P/2)/(10-p)^2 = p^2/(4(10-p))^2`

On taking square root `p^2/(4(10-P))^2 = 1/(1.6 xx  10^5) = (P/2xxP/2)/(10-P)^2 = p^2/4(10-P)^2`

On taking square root  `P^2/(4(10 - P))^2 = (1/(1.6 xx 10^5))^(1/2) or (2(10 - P))/P = (1.6 xx 10^5)^(1/2)`

= `4 xx 10^2`

`20 - 2P = 4xx 10^2 P or P(4xx10^2+2) = 20`

or P = 20/(400 + 2) = 20/ 402 = 0.050 bar

`pH_2 = 0.025 bar; pBr_2 = 0.025 bar: pHBr = 10 - 0.05 = 9.95 bar = 10.0 bar`

Concept: Homogeneous Equlibria - Equilibrium Constant in Gaseous Systems
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 27 | Page 227
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