12. Hydrogen iodide undergoes decomposition according to the equation 2E11(g) -
ID: 698163 • Letter: 1
Question
12. Hydrogen iodide undergoes decomposition according to the equation 2E11(g) - H2(g) + 12(g) The equilibrium constant K, at 500 K for this equilibrium is 0.060. Suppose 0.316 mol of HI is placed in a 8.00-L container at 500 K. What is the equilibrium partial pressure of 1:(g)? (R= 0.0821 L-atm/(K. mol)) a) 0.040 atm b) 0.020 atm c) 0.062 atm d) 0.0078 atm 0.32 atm 13. At 700 K, Kp for the following equilibrium is 5.6 × 10-3. Suppose 77.3 g of mercury(II) oxide is placed in a sealed 1.00-L vessel at 700 K. What is the partial pressure of oxygen gas at equilibrium? (R = 0.0821 L·atm/(K-mol)) a) 10 atm b) 57 atm c) 21 atm d) 0.0056 atm 0.075 atm 14 Consider the following eauilibrium:Explanation / Answer
12)
1st calculate initial pressure of HI using
P*V = n*R*T
P*8.00 = 0.316*0.0821*500
P = 1.621 atm
2HI (g) <---------------> H2 (g) + I2 (g)
1.621 0 0 (initial)
1.621-2x x x (at equilibrium)
Kp = x*x/ (1.621-2x)^2
0.060 = x^2/ (1.621-2x)^2
take square root on both sides,
0.245 = x / (1.621-x)
0.397 - 0.245*x = x
x = 0.32 atm
p(I2) = x = 0.32 atm
Answer: e
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