Answer:
-4
Explanation:
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Let's assume that the gas has ideal gas behavior.
Then we can use ideal gas equation,
PV = nRT
Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)
The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?
By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
n = 11.45 mol
Hence,moles of gas is 11.45 mol.
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Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
