Respuesta :
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
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
Answer:
[tex][HI] _{eq}=0.825mol[/tex]
[tex][H_2] _{eq}=0.010mol[/tex]
[tex][I_2] _{eq}=0.078mol[/tex]
Explanation:
Hello,
At first, the undergoing chemical reaction is:
[tex]H_2(g)+I_2(g)<-->2HI(g)[/tex]
So the law of mass action in terms of concentration turns out into (remember the relationship between Kp and K [tex]Kp=K(RT)^{-\Delta \nu }[/tex]):
[tex]\frac{Kp}{(RT)^{2-2} }=\frac{[HI]^{2}_{eq} }{[H_2]_{eq}[I_2]_{eq}} \\Kp=\frac{[HI]^{2}_{eq} }{[H_2]_{eq}[I_2]_{eq}}[/tex]
Now, we need the initial concentrations which are computed by knowing the volume computed with the ideal gas equation with the total initial moles:
[tex]V=\frac{nRT}{P}=\frac{0.900mol*0.082 \frac{atm*L}{mol*K} *870K}{1bar*\frac{1atm}{1.01325bar} } =65.1L[/tex]
Thus,
[tex][H_2] _0=\frac{0.300mol}{65.1L}=0.005M[/tex]
[tex][I_2] _0=\frac{0.400mol}{65.1L}=0.006M[/tex]
[tex][HI] _0=\frac{0.200mol}{65.1L}=0.003M[/tex]
Thus, by considering the change, we write the law of mass action in terms of the change [tex]x[/tex] due to the equilibrium:
[tex]870=\frac{(0.003+2x)^{2} }{(0.005-x)(0.006-x)}[/tex]
Solving for [tex]x[/tex] we obtain:
[tex]x=0.00484M[/tex]
Thus, the amounts turn out into:
[tex][HI] _{eq} =(0.003M +2(0.00484M))*65.1L=0.825mol[/tex]
[tex][H_2] _{eq} =(0.005M -0.00484M)*65.1L=0.010mol[/tex]
[tex][I_2] _{eq} =(0.006M -0.00484M)*65.1L=0.078mol[/tex]
Best regards.
