Hydrogen iodine decomposes according to the equation 2HI (g) H2 (g) + I2(g) Kc=

Question

Hydrogen iodine decomposes according to the equation 2HI (g) H2 (g) + I2(g)             Kc= 0.0156 at 400 deg C A 0.59 mol sample og HI was injected into a 2.0 L reactionvessel held at 400 deg C. Calculate the concentration of HI atequilibrium Hydrogen iodine decomposes according to the equation 2HI (g) H2 (g) + I2(g)             Kc= 0.0156 at 400 deg C A 0.59 mol sample og HI was injected into a 2.0 L reactionvessel held at 400 deg C. Calculate the concentration of HI atequilibrium

Explanation / Answer

2HI (g)---------------> H2 (g) + I2 (g) 0.59                               0    0 0.59-2x        x    x (0.59-2x)/2                     x/2 x/2 ------------(concentration) Kc=[H2][I2] / [HI]2 =(x/2)(x/2) /[(0.59-2x)/2]2=0.0156 On solving it we get x=0.059 Conc. of HI at equilibrium is (0.59-2x)/2=(0.59 -2 x 0.059)/2=0.236mole / l

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