The heat of combustion of naphthalene C10H8, at constant volume at 25 ...
The heat of combustion at constant volume is given to be -5133 KJ/mol.Therefore, qv=-5133 kJ/mol.The balanced equation of combustion of naphthalene is:C10H8 + 12O2 -----> 10CO2 + 4H2OHere, the number of gaseous moles on the products side= 10 (10 moles of CO2)And, the number of gaseous moles on the reactants side= 12(12 moles of O2)Hence, the change in number of gaseous moles, Δn = 10 - 12 = -2The relation between qp and qv is:qp = qv + ΔnRTSubstituting the values given,qp = -5133*103 + (-2*8.314*298)=> qp= -5137955.144 J/mol
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The heat of combustion of naphthalene C10H8, at constant volume at 25 ...
Delta- * * H =*U + *nRTso * H = - 5133000+[( -2)×8.314×298] thus *H = 5137955.14J or 5137.95515KJ
The heat of combustion of naphthalene C10H8, at constant volume at 25 ...
Enthalpy Change at Constant Pressure
The enthalpy change at constant pressure (ΔH) can be calculated using the heat of combustion (ΔHc) and the difference between the enthalpy change at constant volume (ΔHv) and the gas constant (R).
Formula:
ΔH = ΔHv + RT
Given:
ΔHc = -5133 kJ/mol
Temperature (T) = 25°C = 298 K
R = 8.314 J/(mol·K)
Calculations:
1. Convert the heat of combustion from kJ/mol to J/mol:
ΔHc = -5133 kJ/mol * 1000 J/1 kJ = -5133000 J/mol
2. Calculate the enthalpy change at constant volume (ΔHv):
ΔHv = ΔHc
3. Calculate the enthalpy change at constant pressure (ΔH):
ΔH = ΔHv + RT
ΔH = -5133000 J/mol + (8.314 J/(mol·K) * 298 K)
4. Perform the calculation:
ΔH = -5133000 J/mol + 2479972.872 J/mol
ΔH = -2653027.128 J/mol
Answer:
The enthalpy change at constant pressure (ΔH) for the combustion of naphthalene at 25°C is approximately -2,653,027.128 J/mol.
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