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cricket20 [7]
4 years ago
8

The reaction takes place in water. What happens to the equilibrium when the pressure is increased? a)It favors formation of prod

ucts. b)It is conserved. c)It favors formation of reactants. d)It does not change
Chemistry
1 answer:
OLga [1]4 years ago
6 0

Answer:

I don't really get the options but it favoures the reactant side.

Explanation:

Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g

If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.

With the explanations I have made, I hope the question is now clear to you.

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Consider the reaction:
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<u>Answer:</u> The formation of given amount of oxygen gas results in the absorption of 713 kJ of heat.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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Putting values in above equation, we get:

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2Fe_2O_3\rightarrow 4Fe+3O_2;\Delta H^o_{rxn}=+824.2kJ

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

By Stoichiometry of the reaction:

When 3 moles of oxygen gas is formed, the amount of heat absorbed is 824.2 kJ

So, when 2.594 moles of oxygen gas is formed, the amount of heat absorbed will be = \frac{824.2kJ}{3mol}\times 2.59mol=713kJ

Hence, the formation of given amount of oxygen gas results in the absorption of 713 kJ of heat.

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