ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
Answer:
Check the explanation
Explanation:
Taking a look at the image, I have shown only the energy levels and not the proper curves, because we just need to calculate the energy differences.
Here, S2 denotes a ground state molecule while S2* denotes an excited state molecule
The energy D1, required for dissociation of excited state molecule to 2 ground state atoms is -ve, which means no energy is supplied, rather energy is emitted by this spontaneous dissociation.
Answer:
<em>C</em> C2H5OH = 7.598 molal
Explanation:
- molality (m) C2H5OH = n C2H5OH / Kg H2O
∴ V C2H5OH = 50.0 mL
∴ V H2O = 112.7 mL
∴ T = 20 °C
∴ δ C2H5OH (20°C) = 0.789 g/mL
∴ δ H2O (20°C) = 1.00 g/mL
∴ Mw C2H5OH = 46.0684 g/mol
⇒ n C2H5OH = (50.0 mL)×(0.789 g/mL)×(mol/46.0684 g) = 0.8563 mol
⇒ mass H2O = (112.7 mL)×(1.00 g/mL)×(Kg / 1000 g) = 0.1127 Kg
⇒ <em>C</em> C2H5OH = 0.8563 mol C2H5OH / 0.1127 Kg H2O
⇒ <em>C</em> C2H5OH = 7.598 molal
Answer:
The amount of NaOH required to prepare a solution of 2.5N NaOH.
The molecular mass of NaOH is 40.0g/mol.
Explanation:
Since,
NaOH has only one replaceable -OH group.
So, its acidity is one.
Hence,
The molecular mass of NaOH =its equivalent mass
Normality formula can be written as:
Substitute the given values in this formula to get the mass of NaOH required.
Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g