molar enthalpy of combustion for pentane?
Mass C5H12= 2.15g
Mass H2O= 1.24kg
Initial temp= 18.4°C
Final temp= 37.6°C
Is this in a bomb calorimeter or some other apparatus?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
delta H/g = q/2.15
delta H/mol = (q/2.15)*molar mass pentane
To calculate the molar enthalpy of combustion for pentane, we need to use the data provided and the equation for heat transfer.
First, we need to calculate the amount of heat transferred during the combustion of pentane. The equation for heat transfer is:
q = m × c × ΔT
where:
q = heat transferred (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Given data:
Mass of pentane (C5H12) = 2.15 g
Mass of water (H2O) = 1.24 kg = 1240 g
Initial temperature (water) = 18.4 °C
Final temperature (water) = 37.6 °C
First, let's calculate the heat transferred to water:
q_water = m_water × c_water × ΔT_water
The specific heat capacity of water is approximately 4.18 J/g°C.
q_water = (1240 g) × (4.18 J/g°C) × (37.6 °C - 18.4 °C)
Next, let's calculate the heat transferred from the combustion of pentane:
q_pentane = q_water (since the heat transferred to water equals the heat transferred from pentane)
Finally, let's calculate the molar enthalpy of combustion:
Molar enthalpy of combustion (ΔH_combustion) = q_pentane / moles of pentane
To find the moles of pentane, we need to convert the mass of pentane (C5H12) to moles. The molar mass of pentane can be calculated using the atomic masses of carbon and hydrogen:
Molar mass of pentane (C5H12) = (5 × atomic mass of carbon) + (12 × atomic mass of hydrogen)
Using the atomic masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol), we get:
Molar mass of pentane = (5 × 12.01 g/mol) + (12 × 1.01 g/mol)
Now, we can calculate the moles of pentane using the mass of pentane (2.15 g) and the molar mass of pentane.
Moles of pentane = mass of pentane / molar mass of pentane
Finally, we can calculate the molar enthalpy of combustion by dividing the heat transferred (q_pentane) by the moles of pentane.
I hope this explanation helps you in solving the problem.