Question

The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2HI(9) =H2(g) +129) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.311 MHI. 4.17x10 - MH, and 4.17x10-2 MIL- What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.174 mol of HI(g) is added to the flask? [HI] = [H2) = [12] =

Answer 1

Step-by-step explanation:

The equilibrium constant of the reaction =
`K_c=1.80* 10^(-2)`

Initial concentration of HI = 0.311 M

After addition 0.174 mol of HI(g)

Concentration of HI added =
`(0.174 mol)/(1 L)=0.174 M`

New concentration of HI = 0.311 M + 0.174 M = 0.485 M

`2HI\rightleftharpoons H_2+I_2`

Initial concentration:

0.311 M
`4.71* 10^(-2) M`
`4.71* 10^(-2) M`

At equilibrium:

(0.485 M - x)
`(4.71* 10^(-2) M+x)`
`(4.71* 10^(-2) M+x)`

`K_(eq)=([H_2][I_2])/([HI]^2)`

`1.80* 10^(-2)=((4.71* 10^(-2) M+x)* (4.71* 10^(-2) M+x))/( (0.485 M-x)^2)`

`\sqrt{1.80* 10^(-2)}=((4.71* 10^(-2) M+x))/( (0.485 M-x))`

`0.1342=((4.71* 10^(-2) M+x))/((0.485 M-x))`

`0.065087 M-0.1342x=4.71* 10^(-2) M+x`

`0.065087 M-4.71* 10^(-2) M=1.1342x`

`x=(0.017987 M)/(1.3142)=0.01586 M`

Equilibrium concentrations:

`[HI]=0.485 M-x = 0.485 M - 0.01586 M= 0.46914 M`

`[H_2]=[I_2]=4.71* 10^(-2) M+x=4.71* 10^(-2) M+0.01586 M`

`[H_2]=[I_2]=0.06296 M`