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3 years ago

Given that Delta.G for the reaction below is –957.9 kJ, what is Delta.Gf of H2O?

Chemistry

1 answer:

Alexxandr [17]3 years ago

8 0

Answer:

6ΔG°(f) H₂O = -229 Kj/mol

Explanation:

4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)

ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O

Hess's Law

ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants

-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]

-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj

ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj = -228.56 Kj ≅ -228.6 Kj*

*Verified with Standard Heat of Formation Table

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A certain reaction has an enthalpy of Δ=44 kJ and an activation energy of a=61 kJ. What is the activation energy of the reverse

avanturin [10]

Answer:

E = 17 kJ

Explanation:

The enthalpy of the reaction is:

$\Delta H = E_{p} - E_{r}$

<u>Where</u>:

Ep: is the energy of the products

Er: is the energy of the reactants

Similarly, the enthalpy of the reaction is related to the activation energy forward ( $E_{F}$ ) and to the activation energy reverse ( $E_{R}$ ) as follows:

$\Delta H = E_{F} - E_{R}$

Having that ΔH = 44 kJ and $E_{F}$ = 61 kJ, the activation energy of the reverse reaction is:

$E_{R} = E_{F} - \Delta H = 61 kJ - 44 kJ = 17 kJ$

Therefore, the activation energy of the reverse reaction is 17 kJ.

I hope it helps you!

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3 years ago