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Alisiya [41]
2 years ago
5

Consider the following exothermic reaction and predict how the change below will affect the concentration (increase, decrease, s

tay the same) of each component in the system as the equilibrium is reestablished.
3N2H4 (g) â 4NH3 (g) + N2 (g)

N2 H2 NH3
Add N2(g)
Remove 112(g)
Add NH3(g)
Add Ar(g)
Increase the temperature (constant P)
Decrease the volume (constant T)
Add a catalyst
Chemistry
1 answer:
aev [14]2 years ago
3 0

Answer:

a) equilibrium shifts towards the left

b) equilibrium shifts towards the right hand side.

c)equilibrium shifts towards the left hand side

d) addition of argon has no effect on the equilibrium position

e) equilibrium position shifts towards the left hand side

f) equilibrium position is shifted towards the right hand side

g) addition of a catalyst has no effect on the equilibrium position

Explanation:

Given the equation;

3N2H4(g)⇄ 4NH3 (g) + N2 (g)

Adding N2

The addition of N2 will increase the concentration of N2 in the system thereby shifting the equilibrium position to the left in accordance with Le Chatelier's principle.

Removing N2

The removal of N2 drives the forward reaction and the equilibrium shifts towards the right yielding more N2 in accordance with Le Chateliers principle.

Add NH3

The addition of NH3 will shift the equilibrium position towards the left hand side according to Le Chateliers principle.

d) addition of argon has no effect on the equilibrium position.

e) increasing the temperature

Since the reaction is exothermic, increasing the temperature favours the reverse reaction and the equilibrium position his shifted towards the left hand side.

f) Decrease in volume;

Decreasing the volume favours the forward reaction hence the equilibrium position is shifted towards the right.

g) addition of a catalyst has no effect on the equilibrium position.

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Answer:The sun

Explanation:

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7 0
3 years ago
What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K
vlada-n [284]

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

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Answer:

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Explanation:

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