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natulia [17]
3 years ago
9

How many liters of space will 7.80 moles of methane gas (CH4) occupy at STP

Chemistry
1 answer:
Alchen [17]3 years ago
4 0

Answer:

V CH4(g) = 190.6 L

Explanation:

assuming ideal gas:

  • PV = RTn

∴ STP: T =298 K and P = 1 atm

∴ R = 0.082 atm.L/K.mol

∴ moles (n) = 7.80 mol CH4(g)

∴ Volume CH4(g) = ?

⇒ V = RTn/P

⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)

⇒ V CH4(g) = 190.6 L

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We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

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-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
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K = 9.62


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