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DaniilM [7]
4 years ago
14

Find the ΔG° for the combustion of half a mole of glucose C6H12O6 in kJ

Chemistry
1 answer:
xxTIMURxx [149]4 years ago
8 0

Answer:

-1435 kJ

Explanation:

The Gibbs free energy, ΔG is an indication the amount of (useful) work that can be obtained from a chemical reaction involving reacting species and it is negative for spontaneous reaction

ΔG° = ΔH° - TΔS

ΔG = ΔG° + RT㏑Q

ΔG = G (reactants) - G(products)

Given that the combustion reaction of glucose is as follows;

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

The above reaction has a ΔG° value of -2870 kJ

Therefore, the free energy for half mole of glucose = -2870 kJ/2 = -1435 kJ

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Answer:

E=1.48x10^4J

Explanation:

Hello!

In this case, since the energy implied in a heating process is computed by using the following equation:

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3 0
3 years ago
What is the atomic weight of silver?
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4 years ago
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CHEGG A chemist requires a large amount of 1-bromo-4-phenyl-2-butene as starting material for a synthesis and decides to carry o
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5 0
3 years ago
The pressure in an automobile tire filled with air is 245.0 kPa. If Po2 = 51.3 kPa, Pco2 = 0.10 kPa, and P-others = 2.3 kPa, wha
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Answer:

PN₂ = 191.3 Kpa

Explanation:

Given data:

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6 0
4 years ago
What substance is composed of one type of atom
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