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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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Answer:

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Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

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Mass of the compound = 35.0 mg = 0.035 grams

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We divide by the smallest amount of moes

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N: 0.00152 moles  / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

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