1. what would happen to the position of equilibrium when the following changes are made to the reaction below? 2hg3o (g) ↔ 6hg (g) + o2 (g) δh- (a) hg3o is added to the system (b) the volume of the system decreases (c) temperature is increased 2. balance the exothermic reaction no2 ↔ n2o4 kj – 58.0kj predict the effect of each of the following changes on this system at equilibrium (a) added n2o4 (b) increase the volume (c) add n2 (d) remove no2 (e) decrease the temperature 3. + 333 kj 2no2 ↔ n2 + o2 (a) no2 is added to the system (b) n2 is added to the system (c) o2 is removed from the system (d) the temperature of the container increased (e) the volume of the container increase (f) n2 is added and no2 is removed 4. 179 kj caco3 ↔ cao + co2 (a) co2 is added (b) the volume of the container decreased (c) cao is removed from the system (d) the temperature of the container decreased (e) the volume of the container is increased. (f) caco3 is added to the system 5. h2 + cl2 ↔ 2hcl (a) h2 is removed from the system (b) hcl is removed from the system (c) the volume of the container is removed (d) the temperature of the container is increased (e) the concentration of cl2 is decreased (f) the volume of the container decreased 6. list all the “stressors” that could be applied to this equilibrium reaction, which would cause an increase in the concentration of water vapor. 4hcl (g) + o2 (g) ↔ 2h2o (g) + 2cl2 (g)

Ans. 1) Rxn is

ΔH = - 175.92 KJ/moles

In this rxn the std heat of formation of is 272.1 KJ/mol & std heat of formation of Hg is 61.38 KJ/mol,

So, ΔH = H° (prod) - H° (react) = H° (Hg) - H°

= (6 x 61.38) - (2 x 272.1) = -175.92 KJ/mol.

So, ΔH = -175.92 KJ/mol

Ans. 2) Q.1 a) If is added to the rxn then rxn eqm will shift to -->

If Incr. the conc of the reactants the eqm will shift to --> & produce more products.

As it is an exo rxn it will favor the Fwd rxn.

Ans. 3) Q.1 b) the vol. of the rxn decr. the rxn eqm will shift toward -->.

As the rxn vol. decr. the eqm will shift to the --> as there are more no. of moles present in the rxn.

Ans. 4) Q.1 c) the eqm will shift to -->.

As the temp of the rxn Incr.the eqm will shift toward --> as it is an exo rxn.

Ans. 5) Q.2) The balanced equation is .

2 moles of will give 1 dinitrogen tetraoxide.

Ans. 6) Q.2 a) If more is added to the rxn the rxn eqm will be shifted to <--.

more of will Incr. the prod & so it will shift the eqm to the backward rxn to produce more reactants.

Ans. 7) Q.2 c) if the rxn vol. is decr the eqm will shift to -->.

Decr. in vol. will cause the conc to Incr. & favor the Fwd rxn more.

Ans. 8) Q.2 d) if more N is added to the rxn thenit will have no change in the rxn eqm.

As N is not present in the reactant or product side it won't have any effect on the rxn eqm.

Ans. 9) Q.2 e) if there will be removal of the eqm will shift to <--.

The eqm will change and give more backward rxn to produce .

Ans. 10) Q.2 f) If the rxn temp is decr. the eqm will shift to -->.

As the rxn is exo more amount of product will be generated so the shift will be on Fwd.

Ans. 11) Q.3) balanced rxn.

This rxn has +ve Δ H value so, it is endo rxn.

Ans. 12) Q.3 a) we add the eqm will shift to -->.

more of the reactants will Incr. the product conc by Fwd rxn.

Ans. 13) Q.3 b) If more N is added then the rxn will favor to shift eqm on <--.

Adding of more N will cause an backward rxn resulting in more number of .

Ans. 14) Q.3 c) is removed from the rxn then eqm will shift on -->.

As the number of moles will decr. so more of Fwd rxn will occur.

Ans. 15) Q.3 d) Incr. in temp will result in shifting the eqm to -->.

In endo rxn Incr. in temp will cause more rxn to occur.

Ans. 16) Q.3 e) vol. is Incr then the eqm will shift to <--.

Vol. Incr. will decr. the conc so, the eqm shifts to more no. moles side which is <--.

Ans. 17) Q.3 f) is added & is removed the eqm will shift to <--.

As addition of will cause backward rxn more & removal of will be compensating for addition.

Ans. 18) Q.4) balanced equation. Endo rxn as ΔH is +ve = 179 KJ.

Ans. 19) Q.4 a) is added to the rxn then the eqm shifts to <--.

As more backward rxn will be favored if addition of is done.

Ans. 20) Q.4 b) Vol. decr. will shift the eqm to -->.

As there are more number of moles in the product side the eqm will shift to -->.

Ans. 21) Q.4 c) If CaO is removed then it will shift on -->.

Removal of CaO will produce more CaO than before so the eqm shift will be on -->.

Ans. 22) Q. 4 d) Temp decr. will shift to <--.

As it is endo rxn if temp is decr the eqm will shift to more backward rxn.

Ans. 23) Q. 4 e) Vol. Incr. will shift to -->.

Vol. Incr. will shift the eqm to --> as it has more number if moles in stoichiometry.

Ans. 24) Q.4 f) is added will shift to -->.

As more decomp of will be done.

Ans. 25) Q.5) It is the balanced equation & is exo rxn.

Ans. 26) Q.5 a) If H is removed then shift will be on <--.

As there will be less H more backward rxn will occur in this case.

Ans. 27) Q.5 b) HCl is removed then shift will be on -->.

As eqm will compensate for the removal of HCl it will shift to -->.

Ans. 28) Q.5 c) Vol. is removal is a hypothetical situation.

Ans. 29) Q.5 d) Temp Incr. will shift to -->.

As the rxn is exo the eqm will shift toward -->.

Ans. 30) Q.5 e) Cl is reduced then the rxn will shift to <--.

As cl is reduced the eqm will try to maintain the stable state for chlorine loss & so more backward rxn will occur.

Ans. 31) Q.5 f) If vol. is decr. the shift will be stable.

As both reactants & the products contain same number of moles; So, the eqm will not change on vol. decr..

Ans. 32) Q.6) Stressors will be as follows:

i) Add HCl or molecule.

ii) Incr. the press .of the rxn.

iii) Decr. in temp.

iv) Decr. in vol.

i believe the answer is true.

pv=nrt

explanation: