Respuesta :
Heat energy released is - 31.4 kJ.
Explanation:
CO₍g₎ + 2H₂₍g₎ → CH₃OH₍l₎
Volume of CO, V ₍CO₎ = 15 L = 0.015 m³
Pressure = 112 kPa = 112,000 Pa
T= 85 ⁰C = 85 + 273 = 358 K
As per ideal gas law,
PV = nRT
n = [tex]\frac{PV}{RT}[/tex]
= 112000 × 0.015 / 8.314 × 358
n(CO) = 0.56 moles
Volume of H₂ = 14.4 L = 0.0144 m³
Pressure = 744 torr = 99191.84 Pa
T= 75⁰C + 273 K = 348 K
n(H₂) = 99191.84 ×0.0144 m³ / 8.314 ×348 K
= 0.49 moles of H₂
The above calculation shows that hydrogen is the limiting reagent.
n(CH₃OH) = n(H₂) /2
= 0.49/2 = 0.245 mol
ΔH₍rxn₎ = ΔH₍f₎ (CH₃OH) - ΔH₍f₎(CO)
= -238.6 -(-110.5)
= -128.1 kJ/mol
Now we have to multiply ΔH₍rxn₎ with the moles of methanol.
E = ΔH₍rxn₎ × n(CH₃OH) = -128.1 ×0.245
= - 31.4 kJ.
The reaction produces -4.95 kJ of heat when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr.
The equation of the reaction is;
CO(g) + H2(g) -------> CH2O(g)
The heat of reaction is obtained from;
Enthalpy of products - Enthalpy of reactants = (-116kJ/mol) - (-110.5 kJ/mol)
= -5.5 kJ/mol
Number of moles of CO is obtained from;
PV = nRT
P = 112 kPa or 1.1 atm
T = 85°C + 273 = 358 K
n = ?
R = 0.082 atmLK-1mol-1
V = 15.0 L
n = PV/RT
= 1.1 atm × 15.0 L/ 0.082 atmLK-1mol-1 × 358 K
= 0.56 moles
Number of moles of H2
n = PV/RT
P= 744 torr or 0.98 atm
V = 14.4 L
T = 75°C + 273 = 348 K
n = 0.98 atm × 14.4 L/0.082 atmLK-1mol-1 × 348 K
n = 0.49 moles
We can see that H2 is the limiting reactant here hence 0.49 moles of formaldehyde is produced.
If 1 mole of formaldehyde produces -5.5 kJ of heat
0.49 moles of formaldehyde produces -5.5 kJ × 0.49 moles / 1 mole
= -4.95 kJ of heat
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