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Reil [10]
3 years ago
6

Calculate the free-energy change for the dissolution of 1 mole of nabr and 1 mole of nai at 298 k, given that the values of δh°s

oln are –1 and –7 kj/mol for nabr and nai. the corresponding values of δs°soln are +57 and +74 j/(mol·k).
Chemistry
1 answer:
Georgia [21]3 years ago
7 0
The free energy, which has a symbol of ΔG° at standard conditions has the formula:

ΔG° = n(ΔH° - TΔS°)

For NaBr:
ΔG° = (1 mol)[(-1 kJ/mol)(1000 J/1 kJ) - (298 K)(57 J/mol·K)]
<em>ΔG° = -17,986 J</em>

For NaI:
ΔG° = (1 mol)[(-7 kJ/mol)(1000 J/1 kJ) - (298 K)(74 J/mol·K)]
<em>ΔG° = -29,052 J</em>
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3 years ago
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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

  • dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
  • ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

5 0
4 years ago
How many liters of N2 gas is needed to produce 500 L NH3?
Veseljchak [2.6K]
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6 0
3 years ago
How many atoms are in 6.30 moles of sulfur (S)?
STatiana [176]

Answer: 3.79*10^24 atoms

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8 0
3 years ago
Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

brainly.com/question/13981382

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7 0
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