Answer:
a. 7278 K
b. 4.542 × 10⁻³¹
Explanation:
a.
Let´s consider the following reaction.
N₂(g) + O₂(g) ⇄ 2 NO(g)
The reaction is spontaneous when:
ΔG° < 0 [1]
Let's consider a second relation:
ΔG° = ΔH° - T × ΔS° [2]
Combining [1] and [2],
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)
T > 7278 K
b.
First, we will calculate ΔG° at 25°C + 273.15 = 298 K
ΔG° = ΔH° - T × ΔS°
ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K
ΔG° = 173.1 kJ/mol
We can calculate the equilibrium constant using the following expression.
ΔG° = - R × T × lnK
lnK = - ΔG° / R × T
lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K
K = 4.542 × 10⁻³¹
Lipid you can look up the chemical structures and lipids are the longest which make them the largest
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
