Let a square represent a nitrogen atom and a circle represent an oxygen atom. Co
ID: 485216 • Letter: L
Question
Let a square represent a nitrogen atom and a circle represent an oxygen atom. Consider the exothermic reaction: 2NO(g) N_2(g) + O_2(g) The drawing below shows a small representative volume of a container where the system described by the equation above is initially at equilibrium. a. For each of the following, show a drawing representing the contents of that same small volume after equilibrium is reestablished, and justify your drawing. i. Some gaseous O_2 is removed; temperature and container volume remain constant. ii. Container volume is doubled; temperature remains constant. iii. Temperature is raised by 10 degree C; container volume remains constant. iv. Some gaseous argon is added; temperature and container volume remain constant. b. Can the value for the equilibrium constant for the reaction be determined from the drawing of the small representative volume? If yes, calculate the value for the equilibrium constant. If no, state what additional information you would need, and why you need that information.Explanation / Answer
Le-chatlier's principle:
The factors that affect the equilibrium are changed, the equilibrium will be changed in such a direction which nullify the effect of changed factor.
Case(1): Increase in concentration of reactants or decrease in concentration of products will shift the equilibrium towards right side.
Case(2): Decrease in concentration of reactants or increase in concentration of products will shift the equilibrium towards the left side.
Case(3): Increase in pressure will shift the equilibrium towards less number of gas molecules.
Case(4) Decrease in presssure will shift the equilibrium towards more number of gas molecules.
Case(5): Increase in temperature will shift the equilibrium towards endothermic side.
Case(6) Decrease in temperature will shift the equilibrium towards exothermic side.
2 NO (g) = N2 (g) + O2 (g)
At initial equilibrium,
number of NO molecules = 4
number of N2 molecules = 6
number of O2 molecules = 8
(a)
(i) Removal O2 gas shifts the equilibrium towards right side i.e number of O2 molecules will decreases.
(ii) At initial equilibrium number of both reactant and product gas molecules are same, so the pressure and volume do not affect the equilibrium.
(iii) The forward reaction is endothermic (because deomposition requires energy)
increase in temperature favours the endothermic reaction. So, it shifts the equilibrium towards forward direction.
(iv) Addition of inert gas like argon does not alter the partial pressure of reactaing and product molecules, so it does not affect the equilbirum.
(b)
Equilibrium constant, K = [N2][O2]/[NO]2
K = 6 * 8 / 42
K = 3
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