Answer:
ΔG = - 442.5 KJ/mol
Explanation:
Data Given
delta H = -472 kJ/mol
delta S = -108 J/mol K
So,
delta S = -0.108 J/mol K
delta Gº = ?
Solution:
The answer will be calculated by the following equation for the Gibbs free energy
G = H - TS
Where
G = Gibbs free energy
H = enthalpy of a system (heat
T = temperature
S = entropy
So the change in the Gibbs free energy at constant temperature can be written as
ΔG = ΔH - TΔS . . . . . . (1)
Where
ΔG = Change in Gibb’s free energy
ΔH = Change in enthalpy of a system
ΔS = Change in entropy
if system have standard temperature then
T = 273.15 K
Now,
put values in equation 1
ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)
ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)
ΔG = -472 kJ/mol + 29.5 KJ/mol
ΔG = - 442.5 KJ/mol
Answer:
5.03 moles
Explanation:
Find the molar mass of C5H12 and you will get 72.17 g/mol
Next to find the number of moles, you divide 362.8 by the molar mass and you get
(362.8 g)/(72.17 g/mol)= 5.03 moles
False. That is a compound.
Answer:
I think third C12/hv is right answer
the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
0.00193 mol/L
Given that:
numbers of moles of H₂S = 0.59 moles
Volume = 3.0-L
Equilibrium constant = 9.30 × 10⁻⁸
The equation for the reaction is given as :
2H₂S ⇄ 2H₂(g) + S₂(g)
The initial concentration of H₂S =
The initial concentration of H₂S =
= 0.1966 mol/L
The ICE table is shown be as :
2H₂S ⇄ 2H₂(g) + S₂(g)
Initial 0.9166 0 0
Change -2 x +2 x + x
Equilibrium (0.9166 - 2x) 2x x
(since 2x < 0.1966 if solved through quadratic equation)
The equilibrium concentration for H₂(g) = 2x
∴
= 0.00193 mol/L
Thus, the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
To know more about equilibrium concentration
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