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Sati [7]
3 years ago
5

4 Fe + 6 H₂O + 3 O₂ → 4 Fe(OH)₃

Chemistry
1 answer:
levacccp [35]3 years ago
6 0
Someone answer my question please I need help hurry
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10) At constant temperature, a mixture containing 0.500 M N2 and 0.800 M H2 in a reaction vessel reaches equilibrium. At equilib
jarptica [38.1K]

Answer:

The answer to the question is;

The equilibrium constant for the reaction is 0.278

Reversibility.

Explanation:

Initial concentration = 0.500 M N₂ and 0.800 M H₂

N₂ (g) + 3·H₂ (g) ⇔ 2·NH₃ (g)

One mole of nitrogen combines with three moles of hydrogen form 2 moles of ammonia

That is 1 mole of ammonia requires 3/2 moles of H₂ and 1/2 moles of N₂

0.150 M of ammonia requires 3/2×0.150 moles of H₂ and 1/2×0.150 moles of N₂

That is 0.150 M of ammonia requires 0.225 moles of H₂ and 0.075 moles of N₂

Therefore at equilibrium we have

Number of moles of Nitrogen = 0.500 M - 0.075 M = 0.425 M

Number of moles of Hydrogen = 0.800 M - 0.225 M = 0.575 M

Number of moles of Ammonia = 0.150 M

K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.15)^2}{(0.425)*(0.575)^3}  = 0.278

The kind of reaction is a reversible one as the equilibrium constant is greater than 0.01 which as general guide, all components in a reaction with an equilibrium constant between the ranges of 0.01 and 100 will be present when equilibrium is reached and the chemical reaction will be reversible.

6 0
4 years ago
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