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
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`
