(a) At 800 K the equilibrium constant for I2(g) equilibrium reaction arrow 2 I(g

Question

(a) At 800 K the equilibrium constant for I2(g) equilibrium reaction arrow 2 I(g) is Kc = 3.1 10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.57 10-2 g of I(g), how many grams of I2 are in the mixture?

(b) For 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g), Kp = 3.0 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.31 g of SO3 and 0.189 g of O2. How many grams of SO2 are in the vessel?(b) For 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g), Kp = 3.0 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.31 g of SO3 and 0.189 g of O2. How many grams of SO2 are in the vessel?

Explanation / Answer

moles I = 3.57 x 10^-2 g/ 126.904 g/mol =0.000272 I = 0.000272 mol/ 10.0 L= 0.0000272 M I2 < => 2 I 2.72 x 10^-5 +x .. . . -2x at equilibrium x .. . . . .2.72 x 10^-5 - 2x 3.1 x 10^-5 = ( 2.72 x 10^-5-2x)^2 / x x = [I2] moles I2 = 10.0 x mass I2 = moles I2 x 253.808 g/mol moles SO3 = 1.30 g/80.066 g/mol =0.0162 [SO3] = 0.0162 / 2.00 L =0.00810 M moles O2 = 0.106 g/ 32 g/mol=0.00331 [O2]= 0.00331/ 2.00 L=0.00166 M Kp = Kc (RT)^delta n delta n = 2 - 2 - 1 = -1 3.0 x 10^4 = Kc ( 0.08206 x 700)^-1 = Kc 0.0174 Kc = 1.72 x 10^6 = (0.00810)^2 / [SO2]^2 x 0.00166 = 0.0395/ [SO2]^2 [SO2]= 0.000152 M moles SO2 = 0.000152 x 2.00 L=0.000304 mass SO2 = 0.000304 mol x 64.066 g/mol=0.0195g

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---