Question

Which conditions ( reactant concentration , pressure , and temperature ) elevate high concentration of the underlines in these following equilibrium systems :

1) 2CO + O2 <==> 2CO2 + 167 kJ
2) Cu^+2 + 4NH3 <==> [ Cu(NH3)4 ]^+2 + 24 kJ
3) 2HI + 12.6 kJ <==> H2 + I2
4) 4HCl + O2 <==> 2H2O + 2Cl2 + 113 kJ
5) PCl5 + 88 kJ <==> PCl3 + Cl2

Answer 1

I'm sorry
the underlines are "
1) CO2
2) [ Cu(NH3)4 ]^+2
3) I2
4) 2Cl2
5) Cl2

Answer 2

Here is the first one in detail (for CO2).
1) 2CO + O2 <==> 2CO2 + 167 kJ and I will rewrite the problem to more clearly show how the heat is treated. Basically the heat is made to look just like an added product (if an exothermic rxn) or a reactant (if an endothermic rxn).

1) 2CO + O2 <==> 2CO2 + heat

All of these questions are answered following Le Chatlier's Principle. That says that if we do something to a rxn it will shift so as to undo what we've done to it. Basically we are asking how CO2 is increased by
a. what change in reactant concns(increase or decrease)?
b. what changes in T(increase or decrease)?
c. what changes in pressure(increase or decrease)?

a.If we want to increase CO2 we must increase the concentration of the reactants (put in more CO or O2 or both). That will shift the equilibrium to the right producing more CO2 and more heat. (Adding either reactant makes the reaction try to use up what we've added--that's undoing what we've done-- and that decreases CO and O2 and adds heat and CO2)
b. If we decrease T the reaction will shift so as add heat and that means it will shift to the right. Heat will be added, CO2 will be increased, CO and O2 will decrease.
c. Changes in pressure for gas reactions make the reaction shift to the side with fewer mols for increase in P or to side with more mols for decrease in P. You have 3 mols gas on the left and 2 mol on the right. So increased in P will make it shift to produce more CO2 at the expense of CO and O2.