Question #177071

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H_{2} (*g*) + I_{2} (*g*) ⇌ 2HI (*g*) (400. K for reaction above, K_{p} = 62.5.

(a) 0.40 mol HI are introduced into 1.0 L flask & heated to 400. K, calculate pressure HI (*g*) before reaction occurs.

(b) Find pressure of each species when equilibrium established in (a).

(c) H_{2} (*g*), I_{2} (*g*), HI (*g*) are introduced to flask at 400. K, pressures -->0.40 atm, 0.40 atm, and 5.0 atm respectively. Calculate Q value and directions to reach equilibrium.

(d) H_{2} (*g*), I_{2} (*g*), and HI (*g*) are introduced to flask pressures of 0.040 atm, 0.020 atm, and 0.090 atm, respectively. If partial pressure HI (*g*) decreases 50%, calculate pressure of each species when equilibrium established.

(e) Using (d) carefully draw three curves - one for each- three gases. The curves must show how the pressure of each of the three gases changes as equilibrium is established. Label each curve with the formula of the gas.

Expert's answer

(a) pV = nRT

p x 1.0 L = 0.40 mol x 0.082057 L·atm·K^{-1}·mol^{-1} x 400 K

p = 13.13 atm

(b) K_{p} = p(HI)^{2} / (p(H_{2}) x p(I_{2}))

H_{2} (*g*) + I_{2} (*g*) ⇌ 2HI (*g*)

0 0 13.13 atm start

0+x/2 0+x/2 13.13-x equilibrium

62.5 = (13.13 - x)^{2}/((x/2)(x/2))

7.9 = 2(13.13-x)/x

x = 2.65 atm

(c) Q = p(HI)^{2} / (p(H_{2})·p(I_{2})) = (5.0 atm)^{2} / (0.40 atm x 0.40 atm) = 156.25

Q > K_{p} there are more products than reactants

To decrease the amount of products, the reaction will shift to produce more reactants

(d) p(HI) = 0.090 - 0.045 = 0.045

p(H_{2}) = 0.040 + 0.045/2 = 0.0625

p(I_{2}) = 0.020 + 0.045/2 = 0.0425

(e)

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