What part does chlorophyll play in photosynthesis?
2 answers:
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(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
Answer:
a) ΔHvap=35.3395 kJ/mol
b) Tb=98.62 °C
Explanation:
Given the reaction:
C₇H₁₆ (l) ⇔ C₇H₁₆ (g)
Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.
When
T₁=50°C=323.15K ⇒P₁=0.179
T₂=86°C=359.15K ⇒P₂=0.669
The Clasius-Clapeyron equation is:
ΔHvap=35339.5 J/mol=35.3395 KJ/mol
Normal boiling point ⇒ P=1 atm
Hence, we find the normal boiling point where:
T₁=323.15K
P₁=0.179 atm
P₂=1 atm
T₂=371.77 K= 98.62 °C