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Tasya [4]
3 years ago
15

Predict the shift in equilibrium position that will occur for each of the following reactions when the volume of the reaction co

ntainer is increased.
A) 2COF2(g)⇌CO2(g)+CF4(g).
i) to the left.
ii) to the right.
iii) does not shift.
B) 2NO(g)+O2(g)⇌2NO2(g).
i) to the left.
ii) to the right.
iii) does not shift.
C) 2N2O5(s)⇌4NO2(g)+O2(g).
i) to the left.
ii) to the right.
iii) does not shift.
D) 2SO2(g)+O2(g)⇌2SO3(g).
i) to the left.
ii) to the right.
iii) does not shift.
E) PCl5(g)⇌PCl3(g)+Cl2(g).
i) to the left.
ii) to the right.
iii) does not shift.
Chemistry
1 answer:
enyata [817]3 years ago
5 0

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Increase the volume:

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. So, the equilibrium will shift in a direction where more number gaseous moles are present.

A) 2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

Number of gaseous moles on reactant side = 2

Number of gaseous moles on product side = 2

Equilibrium will not shift any direction as on both sides number of gaseous moles are same.

B) 2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)

Number of gaseous moles on reactant side = 3

Number of gaseous moles on product side = 2

Equilibrium will shift any left direction.

C) 2N_2O_5(g)\rightleftharpoons 4NO_2(g)+O_2

Number of gaseous moles on reactant side = 2

Number of gaseous moles on product side = 5

Equilibrium will shift any right direction.

D) 2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

Number of gaseous moles on reactant side = 3

Number of gaseous moles on product side = 2

Equilibrium will shift any left direction.

E) PCl_5\rightleftharpoons PCl_3(g)+Cl_2(g)

Number of gaseous moles on reactant side = 1

Number of gaseous moles on product side = 2

Equilibrium will shift any right direction.

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Anna007 [38]

The answer to your question is C.)


8 0
3 years ago
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Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

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According to the scientific notation rules, the number to the left of the decimal cannot exceed 10 so we have to move the decimal to the left two spaces. With this change, we also have to change the exponent of the 10. Because we moved the decimal point two spaces to the left, that means we have 10^25.

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