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aleksandrvk [35]
3 years ago
8

What is the ΔG for the following reaction at 25°C?

Chemistry
2 answers:
Pepsi [2]3 years ago
5 0
<span> The ΔG for the following reaction at 25°C of </span><span>N2O4(g) → 2NO2(g) is 5.32 kJ per mole.

2 (51.8) - 98.28 = 5.32 kJ per mole.

This is because t</span>he rate of the dissociation of N2O4 is equal to the rate of formation of N2O4.
ikadub [295]3 years ago
4 0
\Delta G_{global}=\Delta G_{products}-\Delta G_{reagents}\\\\&#10;\text{We should consider the stoichiometry:}\\\\&#10;\Delta G=2\Delta G_{NO_2}-\Delta G_{N_2O_4}\\\\&#10;\Delta G=2\cdot51.8-98.28=103.6-98.28\\\\&#10;\boxed{\Delta G=5.32~kJ}
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Explanation:

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A solution is prepared from 4.5701 g of magnesium chloride and 43.238 g of water. The vapor pressure of water above this solutio
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Answer:

i = 2.483

Explanation:

The vapour pressure lowering formula is:

Pₐ = Xₐ×P⁰ₐ <em>(1)</em>

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ

Where:

Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)

4.5701g of MgCl₂ are:

4.5701g ₓ (1mol / 95.211g) = 0.048000 moles

43.238g of water are:

43.238g ₓ (1mol / 18.015g) = 2.400 moles

Replacing in (1):

0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm

0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)

2.4mol + i*0.048mol = 2.4mol / 0.9527

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i = 0.1192mol / 0.048mol

<em>i = 2.483</em>

<em />

I hope it helps!

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