Given the values of ΔGfo given below in kJ/mol, calculate the value of ΔGo in kJ for the combustion of 1 mole of butane to form
1 answer:
Answer:
Δ
Explanation:
Hello,
Butane's combustion is:
Now the change in the Gibbs free energy for this reaction is computed via (don't forget that oxygen's Gibbs free energy of formation is 0 since it is an element):
Δ4Δ
5Δ
-Δ
Δ
Δ
Now, since the calculation is done for 1 mol of butane, the change in the Gibbs free energy for this fuel is:
Δ
Best regards.
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Answer:
17 ohms
Explanation:
Given that,
In a series circuit there are 17 light bulbs, each providing 1 ohm of resistance.
We need to find the total resistance of the circuit.
The equivalent resistance for series combination is given by :
For 17 light bulbs, the equivalent resistance is given by :
Hence, the total resistance is 17 ohms. Hence, the correct option is (d).
- E(Bonds broken) = 1371 kJ/mol reaction
- E(Bonds formed) = 1852 kJ/mol reaction
- ΔH = -481 kJ/mol.
- The reaction is exothermic.
2 H-H + O=O → 2 H-O-H
There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb
- E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
- ΔH(Breaking bonds) = +1371 kJ/mol
Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release
- E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
- ΔH(Forming bonds) = - 1852 kJ/mol
Heat of the reaction:
is negative. As a result, the reaction is exothermic.