the combustion of co2 gives -393.5kj/mol calculate the heat of formation of 35.2grams of co2
enthalpy of combustion of CO2=-393.5 kJ/mol
enthalpy of formation of CO2 = -enthalpy of combustion of CO2 = -(-393.5 kJ/mol) = 393.5 kJ/mol
number of moles of CO2 = mass CO2/molecular mass of CO2 = 35.2 g/ (12.01+16.0x2)g/mol =0.800 mol
1 mol of CO2 formed using 393.5 kJ
0.800 mol of CO2 formed using 0.8x393.5 kJ = 314.8 kJ
How does one combust CO2? Burn it with oxygen? fat chance.
Thanks this answer is absolutely correct
To calculate the heat of formation of CO2, we need to use the equation:
ΔHf = ΔHc/n
Where:
ΔHf = Heat of formation
ΔHc = Heat of combustion
n = Number of moles
First, we need to calculate the number of moles of CO2. We can use the molar mass of CO2 to do this:
Molar mass of CO2 = Atomic mass of carbon + 2 * atomic mass of oxygen
= 12.01 g/mol + 2 * 16.00 g/mol
= 44.01 g/mol
Now, we can calculate the number of moles using the given mass:
Number of moles = Mass / Molar mass
= 35.2 g / 44.01 g/mol
Next, we can substitute the values into the equation to find the heat of formation:
ΔHf = (-393.5 kJ/mol) / (35.2 g / 44.01 g/mol)
Now, solve the equation:
ΔHf = -393.5 kJ/mol * (44.01 g/mol / 35.2 g)
Finally, calculate the heat of formation:
ΔHf ≈ -494.98 kJ/mol
Therefore, the heat of formation of 35.2 grams of CO2 is approximately -494.98 kJ/mol.