Hydrogen iodide decomposes according to the followingequation 2HI(g) ---> H2(g)

Question

Hydrogen iodide decomposes according to the followingequation 2HI(g) ---> H2(g) + I2(g)    Kc=.0156@400deg celcius Initially, 0.550 mol HI was injected into an empty 2.0Lreaction vessel at 400deg cel. Calculate the concentration of HI atequilibrium? Hydrogen iodide decomposes according to the followingequation 2HI(g) ---> H2(g) + I2(g)    Kc=.0156@400deg celcius Initially, 0.550 mol HI was injected into an empty 2.0Lreaction vessel at 400deg cel. Calculate the concentration of HI atequilibrium?

Explanation / Answer

                                       2HI(g)    --->       H2(g)   +     I2(g)   initialconc.                   0.550 /2                  0                  0 Equb.conc.                  0.275 -x                  x                 x Where x is the equilibrium concentration ofproducts equilibrium constant K c = [ H 2 ] [ I 2 ] /[HI]^ 2                           0.156 = x(x ) / (0.275-x) ^2                              x/ ( 0.275-x ) = 0.156                                      =0.3949                                  x = 0.3949 ( 0.275-x)                                  x = 0.0778 moles / L Therefore equilibrium concentration of HI = 0.275-0.0778                                                               = 0.1971 moles/L

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