An equilibrium mixture of H2, I2, and HI at 458 C contains 0.112 molH2, 0.112 mo
ID: 554171 • Letter: A
Question
An equilibrium mixture of H2, I2, and HI at 458 C contains 0.112 molH2, 0.112 molI2, and 0.775 molHI in a 5.00-L vessel.
1)What are the equilibrium partial pressure of
HI when equilibrium is reestablished following the addition of 0.200 mol of HI?
Express your answer to four significant figures and include the appropriate units
2)What are the equilibrium partial pressure of
I2 when equilibrium is reestablished following the addition of 0.200 mol of HI?
Express your answer to three significant figures and include the appropriate units.
3) What are the equilibrium partial pressure of
H2 when equilibrium is reestablished following the addition of 0.200 mol of HI?
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
H2(g) + I2(g) <---> 2HI(g)
at equilibrium,
concentration of H2 = 0.112/5 = 0.0224 M
concentration of I2 = 0.112/5 = 0.0224 M
concentration of HI = 0.775/5 = 0.155 M
Kc = 0.155^2/(0.0224*0.0224) = 47.88
concentration of HI added = 0.2/5 = 0.04 M
total concentration of HI = 0.155+0.04 = 0.195 M
47.88 = (0.195-2x)/(0.0224+x)^2
x = 0.0305
1)concentration of HI after reestablishing equilibrium
= 0.195-(2*0.0305)
= 0.134 M
partial pressure of HI = nRT/V
= 0.134*0.0821*(458+273.15)
= 8.044 atm
2) concentration of I2 after reestablishing equilibrium
= 0.0224+0.0305
= 0.0529 M
partial pressure of I2 = nRT/V
= 0.0529*0.0821*(458+273.15)
= 3.17 atm
3) concentration of H2 after reestablishing equilibrium
= 0.0224+0.0305
= 0.0529 M
partial pressure of I2 = nRT/V
= 0.0529*0.0821*(458+273.15)
= 3.17 atm