Question

A sealed container was filled with 0.300 mol h2(g), 0.400 mol i2 (g), and 0.200 mol hi (g) at 870k and total pressure 1.00 bar. calculate the amounts of the components in the mixture at equilibrium given that k = 870 for the reaction h2 (g) + i2 (g) ⇌ 2 hi (g).

Answer 1

`[HI] _(eq)=0.825mol`

`[H_2] _(eq)=0.010mol`

`[I_2] _(eq)=0.078mol`

Step-by-step explanation:

Hello,

At first, the undergoing chemical reaction is:

`H_2(g)+I_2(g)<-->2HI(g)`

So the law of mass action in terms of concentration turns out into (remember the relationship between Kp and K
`Kp=K(RT)^(-\Delta \\u  )`):

`(Kp)/((RT)^(2-2) )=([HI]^(2)_(eq) )/([H_2]_(eq)[I_2]_(eq))  \\Kp=([HI]^(2)_(eq) )/([H_2]_(eq)[I_2]_(eq))`

Now, we need the initial concentrations which are computed by knowing the volume computed with the ideal gas equation with the total initial moles:

`V=(nRT)/(P)=(0.900mol*0.082 (atm*L)/(mol*K) *870K)/(1bar*(1atm)/(1.01325bar) )  =65.1L`

Thus,

`[H_2] _0=(0.300mol)/(65.1L)=0.005M`

`[I_2] _0=(0.400mol)/(65.1L)=0.006M`

`[HI] _0=(0.200mol)/(65.1L)=0.003M`

Thus, by considering the change, we write the law of mass action in terms of the change
`x` due to the equilibrium:

`870=((0.003+2x)^(2) )/((0.005-x)(0.006-x))`

Solving for
`x` we obtain:

`x=0.00484M`

Thus, the amounts turn out into:

`[HI] _(eq) =(0.003M +2(0.00484M))*65.1L=0.825mol`

`[H_2] _(eq) =(0.005M -0.00484M)*65.1L=0.010mol`

`[I_2] _(eq) =(0.006M -0.00484M)*65.1L=0.078mol`

Best regards.

Answer 2

Given:

Moles of H2 = 0.300

Moles of I2 = 0.400

Moles of HI = 0.200

Keq = 870

To determine:

Amounts of the mixture at equilibrium

Step-by-step explanation:

H2(g) + I2(g) ↔ 2HI(g)

Initial 0.3 0.4 0.2

Change -x -x +2x

Eq (0.3-x) (0.4-x) (0.2+2x)

Keq = [HI]²/[H2][I2]

870 = (0.2+2x)²/(0.3-x)(0.4-x)

x = 0.29 moles

Amounts at equilibrium:

[HI] = 0.2 + 2(0.29) = 0.78 moles

[H2] = 0.3-0.29 = 0.01 moles

[I2] = 0.4-0.29 = 0.11 moles