Question

help with questions please

TA'[B Write the expression for the equilibrium constant for the following reactions: N204 (g) 2NO2 (g) N2 (9) + 3H2 (g) = 2NH3 (9) Solids are not included in the expression since the concentrations do not change. After you write the expression, explain why the solid concentrations do not change. MgCO3(s) = MgO(s) + CO2(g) Solvents are not included in the expression since the concentrations do not change. After you write the expression, explain why the solid concentrations do not change. NH: (aq) + H2OU NH4+ (aq) + OH(aq)

Answer 1

Ke=CNo [N2O4] kc=N

Solid and liquids are not included in expression because their effective concentrations stay constant throughout the reaction

ke ke = [ouN,

Peniy 0- 715 C2) И = 0.324 QrKp equilibrum Not at direction, Lo ttut k reached COm re Ke= kp(er) -(-2) .(0 0b21x C50) Kc 2.7 (0.021 X 650) Kc 144.08

ССHU КС 4.08 Сv] o 0635M 144.08

e)

Lechalteir principle says any change in equilibrium conditions will shift the equilibrium in that direction which will resist the change.

Factors

a) Increasing/ Decreasing the concentration of reactants or products:

If a reactant concentration is increased by adding more of it, the equilibrium will move to forward direction so that concentration of increased reactant can be decreased.

If a product concentration is increased by adding, the equilibrium will shift to backward direction so that concentration of increased product can be decreased.

If a reactant concentration is decreased by adding a reagent which can react with that reactant, the equilibrium will shift to backward direction so that concentration of decreased reactant can be increased.

If a product concentration is decreased by adding a reagent which can react with that product, the equilibrium will shift to forward direction so that concentration of decreased product can be increased.

b) Increase/ decrease in temperature

case I) If the reaction is exothermic in nature means heat releasing.

if we increase the temperature, the equilibrium will shift backwards to reduce the temperature of the system.

if we decrease the temperature, the equilibrium will shift forward to increase the temperature of the system.

case II) If the reaction is endothermic in nature means heat absorbing.

if we increase the temperature, the equilibrium will shift forward to use the temperature or heat given to the system.

if we decrease the temperature, the equilibrium will shift backward to increase the temperature of the system.

c) Increase / decrease in pressure:

If we increase the pressure equilibrium will shift to that side which has less number of moles.

If we decrease the pressure equilibrium will shift to that side which has more number of moles.

now lets solve the question

I) adding carbon: more C on the reactant side is added, but theconcentration of the C doesn't change. Therefore, it has no effect on the position of equilibrium.

II) If volume of vessel is decreased, pressure is increased. This will shift equilibrium to that side which has less number of moles.That means to the product side in forward direction as it has 1 mole of methane gas whereas reactant side has 2 moles.

III) If hydrogen is removed, Concentration of Reactant is decreased.It will shift the equilibrium to backward direction so that concentration of decreased reactant can be increased.

IV) A catalyst affect the speed of the two reactions in equal amount but has no effect on rate of reaction. Hence no effect on equilibrium.

kcrevee Kc 14u.00 7000614