The heat of combustion of C2H4 at 27 C at constant prssure is -333.3 Kcal. What will be the heat of combustion at constant volume considering the water to be in liquid state? (R=1.987 cal)
To calculate the heat of combustion at constant volume, we need to consider the difference in enthalpy between the initial and final states. In this case, we'll consider the water to be in the liquid state.
The heat of combustion at constant pressure (ΔH) is given as -333.3 Kcal. But we need to convert this value into calories because the value of R (gas constant) is given in cal.
1 Kcal = 1000 cal, so -333.3 Kcal = -333,300 cal.
Now, to calculate the heat of combustion at constant volume (ΔU), we need to use the following equation:
ΔH = ΔU + ΔnRT
Where:
ΔH = heat of combustion at constant pressure
ΔU = heat of combustion at constant volume
Δn = change in the number of moles of gas (reactants - products)
R = gas constant
T = temperature in Kelvin
In this case, the reaction is the combustion of C2H4, which produces CO2 and H2O. The balanced chemical equation is:
C2H4 + O2 -> CO2 + H2O
From the balanced equation, we can see that Δn = 0 because the number of gas molecules remains the same before and after the reaction.
So the equation becomes:
ΔH = ΔU + 0
Therefore:
ΔU = ΔH
Substituting the value of ΔH = -333,300 cal, we get:
ΔU = -333,300 cal
Hence, the heat of combustion at constant volume considering the water in the liquid state is -333,300 cal.