Calculate the enthalpy change in kilojoules when 54.7g of MgCO3 decomposes according to the following equation:
MgCO3 (s) into MgO (s) + CO2 (g)
My delta H answer was 100.6 KJ but the real answer is 76.1 KJ. Can someone please help?
I don't have MgCO3 delta H.
I have -393.5 kJ for CO2 and -601.8 kJ for MgO. I found -1112 kJ on the web for MgCO3 but I don't know how reliable that is. If we take that number, then
delta Hrxn = (products)-(reactants) =
(MgO)+(CO2)-(MgCO3)
(-601.8) + (-393.5) -(-1112) = 116.7 kJ/mol MgCO3.
We have 54.7 grams = 54.7/84.31 = 0.649 moles. Then 116.7 x 0.649 = 75.7 kJ
which is close. You must have them listed in your text or in the problem Just substitute the numbers you have.
To calculate the enthalpy change during the decomposition of MgCO3, we need to use the balanced equation and the molar masses of the compounds involved.
The balanced equation for the decomposition of MgCO3 is:
MgCO3(s) -> MgO(s) + CO2(g)
First, let's calculate the molar mass of MgCO3:
- Mg = 24.31 g/mol
- C = 12.01 g/mol
- O = 16.00 g/mol
Molar mass of MgCO3 = 24.31 + 12.01 + (3 * 16.00) = 84.31 g/mol
Next, let's calculate the number of moles of MgCO3:
Number of moles = Mass / Molar mass
Number of moles = 54.7 g / 84.31 g/mol = 0.649 mol
Now, we need to use the stoichiometric coefficients from the balanced equation to determine the molar ratio for the enthalpy change:
From the balanced equation: 1 mol of MgCO3 -> 1 mol of MgO
So, the number of moles of MgO formed will be equal to 0.649 mol.
Next, we need to calculate the enthalpy change using the enthalpy of formation values.
The enthalpy change for the decomposition of 1 mol of MgCO3 is given as -1171 kJ/mol.
Enthalpy change = -1171 kJ/mol * 0.649 mol = -759.679 kJ (rounded to three decimal places)
Therefore, the enthalpy change during the decomposition of 54.7g of MgCO3 is approximately -759.679 kJ.
Since the actual answer you mentioned is 76.1 kJ, please double-check the enthalpy change value given in the question or provide any additional information if available.
To calculate the enthalpy change during the decomposition of MgCO3, you need to use the concept of heat of formation and Hess's law. The enthalpy change can be determined by the difference in the heat of formation between the reactants and products.
First, let's find the heat of formation for the reactant MgCO3 and products MgO and CO2. The heat of formation values can be found in reference tables or online resources.
The heat of formation of MgCO3 is given as -1111 kJ/mol.
The heat of formation of MgO is given as -601 kJ/mol.
The heat of formation of CO2 is given as -393.5 kJ/mol.
Next, we need to determine the stoichiometric coefficients for the balanced chemical equation. From the balanced equation:
MgCO3 (s) -> MgO (s) + CO2 (g)
It can be observed that one mole of MgCO3 decomposes to give one mole of MgO and one mole of CO2.
Now we can calculate the enthalpy change using the equation:
ΔH = (Σn x ΔHf)products - (Σn x ΔHf)reactants
Where ΔHf is the heat of formation and 'n' is the stoichiometric coefficient.
For the products, we have:
ΔH = [(1 x ΔHf)MgO + (1 x ΔHf)CO2] - [(1 x ΔHf)MgCO3]
Substituting the values:
ΔH = [(1 x -601 kJ/mol) + (1 x -393.5 kJ/mol)] - [(1 x -1111 kJ/mol)]
ΔH = [-601 kJ/mol - 393.5 kJ/mol] - [-1111 kJ/mol]
ΔH = -994.5 kJ/mol - (-1111 kJ/mol)
ΔH = -994.5 kJ/mol + 1111 kJ/mol
ΔH = 116.5 kJ/mol
Finally, convert the enthalpy change from kJ/mol to kJ/g by dividing by the molar mass of MgCO3.
The molar mass of MgCO3 = 24.31 g/mol + 12.01 g/mol + (3 x 16.00 g/mol) = 84.31 g/mol
ΔH = (116.5 kJ/mol) / 84.31 g/mol
ΔH = 1.38 kJ/g
Now, calculate the enthalpy change for the given mass of MgCO3 (54.7 g) by multiplying it with the enthalpy change per gram:
ΔH = (1.38 kJ/g) x (54.7 g)
ΔH = 75.526 kJ
Therefore, the enthalpy change during the decomposition of 54.7 g of MgCO3 is approximately 75.526 kJ, which is close to the given answer of 76.1 kJ.