Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
Answer:
5.35m H2O2 x 34.02g/1m H2O2 = 182g H2O2
Explanation:
Answer:
e
Explanation:
<em>Provided the reaction that leads to the formation of the products can proceed in both forward and backward directions, the correct answer would be yes because the reaction will proceed backward until equilibrium is reached.</em>
<u>For a reaction that can proceed both forward and backward, the addition of a catalyst increases the rate of reaction in both directions based on the fact that a catalyst cannot alter the equilibrium of a reaction. </u>
Hence, if an enzyme is added to the product of a reaction that has the potential to proceed in both forward and reverse reactions, a substrate would be expected to form because the reaction will proceed backward until an equilibrium is reached.
The correct option is e.
Answer:
Explanation:
AgCl ⇄ Ag⁺ + Cl⁻
m m m
If x mole of AgCl be dissolved in one litre .
[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰
m² = 1.6 x 10⁻¹⁰
m = 1.26 x 10⁻⁵ moles
So solubility of AgCl is 1.26 x 10⁻⁵ moles / L
