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Shalnov [3]
3 years ago
14

Consider the following system at equilibrium:

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

Increase D  ⇒ rightward shift

Increase E  ⇒ rightward shift

Increase F  ⇒ leftward shift

Decrease D  ⇒ leftward shift

Decrease E  ⇒ leftward shift

Decrease F ⇒ rightward shift

Triple D and reduce E to one third  ⇒ no shift

Triple both E and F ⇒ no shift

Explanation:

Changes in the concentrations of reactants y products affect to the equilibrium. In this case, D and E are reactants and F is the product.

When a change in the factors affecting equilibrium occur, the system will react in order to counteract the change imposed.

  • If a increase of reactant occur, the system will tend to produce more products (<em>rightward shift</em>). For example: increase D, increase E. In contraposition, if a decrease of reactant occur, the system will produce more reactants (<em>leftward shift</em>). For example: decrease D, decrease E.
  • If a increase of product occur, the system will tend to produce more reactants (<em>leftward shift</em>). For example: increase F. In contraposition, if a decrease of product occur, the system will produce more products (<em>rightward shift</em>). For example: decrease F.
  • If we change both reactants and products concentration in the same amount, the net change is zero, so there is <em>not shift</em>. For example: Triple D and reduce E to one third (D x 3 and E/3, there is not change in the total amount of reactants) ; Triple both E and F (E x 3 and F x 3, the net change is zero).
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Given: Fe2O3(s) + 3 CO(g) –––––&gt; 2 Fe(s) + 3 CO2(g); ΔH° = –26.8 kJ FeO(s) + CO(g) –––––&gt; Fe(s) + CO2(g); ΔH° = –16.5 kJ d
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ΔH° = + 6.2 kJ

Explanation:

Fe₂O₃(s) + 3CO(g) –––––> 2 Fe(s) + 3 CO₂(g);        ΔH° = –26.8 kJ  ......... ( 1 )

  FeO(s) + CO(g) –––––> Fe(s) + CO₂(g);                  ΔH° = –16.5 kJ  .........( 2 )

   Multiplying equation ( 2 ) by 2 and subtracting it from ( 1 )

  Fe₂O₃(s) + 3CO(g) -2 FeO(s) -  2CO(g) ––> 2 Fe(s) + 3 CO₂(g) - 2Fe(s) -           2CO₂(g)              ΔH° = –26.8 kJ  -  ( 2 x –16.5 kJ )

Fe₂O₃(s) + CO(g) -2 FeO(s)––>  CO₂(g)              ΔH° = –26.8 kJ + 33 kJ

Fe₂O₃(s) + CO(g) ––>2 FeO(s) +CO₂(g)               ΔH° = + 6.2 kJ

Required   ΔH° = + 6.2 kJ       Ans .

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