Homework Assignment #4 (Due date: October 30, 2017) (1) Consider the following e

Question

Homework Assignment #4 (Due date: October 30, 2017) (1) Consider the following equilibrium process at 700 C Analysis shows that there are 2.50 moles of Ha, 1.35 10 mole of Sa, and 8.70 moles of HsS present in a 12.0 L nask. Calculate the equilibrium, constant K, for the reaction. (2) The equitibrium constant Ke for the reaction the is 3,8 x 10- at 727 C. Calculate Ke and Kp for the equilibrium at the same temperature. (3) For the reaction equilibrium if a mixture of 0.400 at 700°C. A, 0.534. Calculate the number of moles of 12 that are present at equi mole of CO and 0.400 mole of H:O is heated to 700°C in a 10.0-L container (4) The equilibrium constant Ke for the reaction is 4.2 at 1650 °C. Initially 0.80 mol Ha and 0.80 mol CO; are injected into a 5.0-L flask. Calculate the concentration of each species at equilibrium. (5) Consider the reaction /r "-198.2 kJ/mol Comment on the changes in the concentrations of SO:, O, and SO, at equilibrium if we were to (a) increase temperature; (b) increase the pressure; (c) increase SO2; (d) add a catalyst; (e) add helium at constant volum (6) Consider the equilibrium system 3AB. Sketch the changes in the concentrations of A and B over for the following situations: (a) initially only A is present; (b) initially only B is present; (c) initially both A B are present (with A in higher concentration). In each case, assume that the concentration of B is higher that of A at equilibrium.

Explanation / Answer

4)

initially,

[H2] = mol of H2 / volume in L

= 0.80 mol/ 5.0 L

= 0.16 M

[CO2] = mol of CO2 / volume in L

= 0.80 mol/ 5.0 L

= 0.16 M

[H2] [CO2] [H2O] [CO]

initial 0.16 0.16 0 0

change -1x -1x +1x +1x

equilibrium 0.16-1x 0.16-1x +1x +1x

Equilibrium constant expression is

Kc = [H2O]*[CO]/[H2]*[CO2]

4.2 = (1*x)^2/(0.16-1*x)^2

sqrt(4.2) = (1*x)/(0.16-1*x)

2.04939015319192 = (1*x)/(0.16-1*x)

0.3279-2.04939*x = 1*x

0.3279-3.04939*x = 0

x = 0.108

At equilibrium:

[H2] = 0.16-x = 0.16-0.108 = 0.052 M

[CO2] = 0.16-x = 0.16-0.108 = 0.052 M

[H2O] = x =0.108 = 0.108 M

[CO] = x = 0.108 = 0.108 M

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