Question

Part A:

Consider the following system at equilibrium: A(aq)+B(aq)⇌2C(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.

Bins: leftward shift, rightward shift, no shift

Categories: Increase [B], Decrease [C], Decrease [B], Decrease [A], Double [A] and halve [B], Double both [B] and [C], Increase [A], increase [C]

Part B:

The following system is at equilibrium:2X(s)+4Y(g)⇌Z(g)Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction.

Bins: leftward shift, rightward shift, no shift

Categories: Halve the volume, Remove some X, Add more X, Double the volume

Answer 1

1)According to Le-Chatlier principle equilibrium will try to reach its normal position when something is added or removed

when any reactant is added,equilibrium shifts in forward direction

when any reactant is removed,,equilibrium shifts backward

when any product is added,equilibrium shifts in backward direction

when any product is removed,,equilibrium shifts forward

Note - concentration of solids,H₂O doesnt effects equilibrium

A(aq)+B(aq)⇌2C(aq) K_eq = [C]²/[A][B]

Increase [B] - increase in reactant, equilibrium shifts in forward direction

Decrease [C] - decrease in product, equilibrium shifts in backward direction

Decrease [B] - decrease in reactant, equilibrium shifts in backward direction

Decrease [A] - decrease in reactant, equilibrium shifts in backward direction

Double [A] and halve [B] - since [A] and [B] are in same ration so change in concentration will be neutralised,no shift

Double both [B] and [C] - in K_eq if we double C and half B,equilibrium will increase ,so to bring back value again at K_eq, [A] must increase,so reaction in backward direction

Increase [A]- increase in reactant, equilibrium shifts in forward direction

increase [C]-increase in product, equilibrium shifts in backward direction

2) according to le chatlier principle

if volume is increased, equilibrium shifts in direction which has more moles of gas

if volume is decreased, equilibrium shifts in direction which has few moles of gas

2X(s) +4Y(g)⇌Z(g)

K_eq - [Z] / [Y]⁴

concentration of X will not effect equilibrium as it is solid

Halve the volume- Volume decreases, product side has 1 mole of gas while reactant side has 4 moles, so reaction will shift forward

Remove some X- No shift as X is solid

Add more X - No shift as X is solid

Double the volume- Volume increases, product side has 1 mole of gas while reactant side has 4 moles, so reaction will shift backward