1. At 323*C, there are 0.10 moles H 2 (g), 0.20 moles HClO 4 (g), 0.10 moles H2O
ID: 993624 • Letter: 1
Question
1. At 323*C, there are 0.10 moles H2 (g), 0.20 moles HClO4 (g), 0.10 moles H2O (g), and 0.36 moles HCl (g) at equilibrium in a 400-mL flask equilibrium mixture. What is the value of Keq for this reaction?
1 HClO4 (g) + 4H2 (g) <—-> 1HCl (g) + 4H2O (g)
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2. For the following reaction, 0.20 moles Bi2S3 (s) are mixed with 0.50 moles H2 (g). Once equilibrium is established, 0.225 moles of H2 (g) remain.
What is the value of Keq for this reaction?
1 Bi2S3 (s) + 3H2 (g) <——> 2Bi (s) + 3 H2S (g)
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3. At 92.2*C, the Kp for the following reaction is 0.2000 atm-1. If you were to place exactly 0.200 atm of N2O4 (g) into a 1.00 liter vessel, what would the partial pressure of NO2 (g) be once equilibrium was established?
2NO2 (g) <——> N2O4 (g)
Explanation / Answer
in equilibrium
K = [HCl][H2O]^4 / ([HClO][H2]^4)
[HCl] = 0.36/0.4 = 0.9
[H2O] = 0.1/0.4 = 0.25
[HClO] = 0.2/0.4 = 0.5
[H2] = 0.1/0.4 = 0.25
K = [HCl][H2O]^4 / ([HClO][H2]^4) = (0.9)(0.25^4) / ((0.5)(0.25^4))
K = 1.8
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