Answer to Question #177071 in General Chemistry for dsa

Question

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H₂ (g) + I₂ (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) H₂ (g), I₂ (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) H₂ (g), I₂ (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. 

Answer 1

(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(H₂) x p(I₂))

H₂ (g) + I₂ (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(H₂)·p(I₂)) = (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(H₂) = 0.040 + 0.045/2 = 0.0625

p(I₂) = 0.020 + 0.045/2 = 0.0425

(e)