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konstantin123 [22]
3 years ago
8

Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘c.

Chemistry
1 answer:
gayaneshka [121]3 years ago
4 0

This is an incomplete question, here is a complete question.

Consider the reaction between nitrogen and oxygen gas from dinitrogen monoxide.

2N_2(g)+O_2(g)\rightarrow 2N_2O(g)

Given: delta Hrxn= +163.2kJ

Calculate the entropy change in the surroundings when this reaction occurs at 25 degrees C.

Answer : The entropy change in the surroundings is, -547.6 J/K

Explanation :

Formula used to calculate the entropy change in the surroundings is:

\Delta S_{surr}=-\frac{Q}{T}

where,

\Delta S_{surr} = entropy change in the surrounding

Q = heat energy

T = temperature = 25^oC=273+25=298K

Given:

\Delta H_{rxn}=+163.2kJ

\Delta H_{rxn}=Q=+163.2kJ

Now put all the given values in the above formula, we get:

\Delta S_{surr}=-\frac{163.2kJ}{298K}

\Delta S_{surr}=-\frac{163.2\times 1000J}{298K}

\Delta S_{surr}=-547.6J/K

Therefore, the entropy change in the surroundings is, -547.6 J/K

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Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

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  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of CuFeS_2 = 866.337 g = 0.8663 Kg         (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.


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