iglumelptsRecovalentActivity.do?locatorassignment-take okmarks here on the bookm

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iglumelptsRecovalentActivity.do?locatorassignment-take okmarks here on the bookmarks bar. Import bookmarks now... Use the References to access important values if needed for this question. The equilibrium constant, Ko, for the following reaction is 55.6 at 698 K H2(8)+128)H Calculate the equilibrium concentrations of reactants and product when 0.328 moles of H, and 0.328 moles of ly are introduced into a 1.00L vessel at 698 K [12] = HI] Submit Answer Retry Entire Group 9 more group attempts remaining 120

Explanation / Answer

1)

          H2(g) + I2(g) ------------------------ 2 HI(g)                 Kc= 55.6

       0.328      0.328                                   0

        -x             -x                                       +2x

      0.328-x   0.328-x                                 +2x

Kc = [HI]^2/[H2][I2]

55.6 = (2x)^2/(0.328-x)(0.328-x)

55.6 = 4x^2/(0.328-x)^2

for solving the equation

x= 0.259

at equilibrium

number of moles of H2= 0.328-x=0.328-0.259 =0.069 moles

number of moles of I2=0.328-x=0.328-0.259 =0.069 moles

number of moles ofHI= 2x=2x0.259=0.518 moles

Volume= 1L

[H2]=numver of moles/volume = 0.069/1=0.069M

[I2]=0.069/1=0.069M

[HI]= 0.518/1= 0.518M

2)

    NH4HS(s)   -------------- NH3(g)    + H2S(g)            Kc=1.80x10^-4

0.593                                  0               0

-x                                     +x               +x

0.593-x                              +x               +x

Kc= [NH3][H2S]/[NH4HS]

1.80x10^-4 = x*x/( 0.593-x)

for solving the equation

x= 0.010

at equilibrium

number of moles of H2S= 0.010 moles

Volume = 1.0L

Concentration of H2S = 0.010/1.0= 0.010M

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