A scientist measures the standard enthalpy change for the following reaction to be -2923.0 kJ :
2C2H6(g) + 7 O2(g) 4CO2(g) + 6 H2O(g)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CO2(g) is _________
dHrxn = (n*dHf products( - (n*dHf reactants)
2923.0 kJ = (4*dHCO2 + 6*dHH2O) - (2*dHC2H6). Look up dH H2O and dH C2H6, substitute and solve for dH CO2.
Horxn = (4 mol)Hf(CO2(g)) + (6 mol)Hf(H2O(g)) - [(2 mol)Hf(C2H6(g)) + (7 mol)Hf(O2(g))] = -2909.8 kJ
Solve for Hf(CO2(g)) in the above equation:
Hf(CO2(g)) = [Horxn - (6 mol)Hf(H2O(g)) + (2 mol)Hf(C2H6(g)) + (7 mol)Hf(O2(g))] / (4 mol)
= [(-2909.8 kJ) - (6 mol)(-241.8 kJ/mol) + (2 mol)(-84.7 kJ/mol) + (7 mol)(0.0 kJ/mol)]/(4 mol)
=-407.1 kJ/mol
To find the standard enthalpy of formation of CO2(g) using the given standard enthalpy change and the standard enthalpies of formation for the other substances, we can use the following equation:
ΔH°rxn = ΣnΔH°f(products) - ΣmΔH°f(reactants)
Where:
ΔH°rxn is the standard enthalpy change for the reaction
ΔH°f is the standard enthalpy of formation
n and m are the stoichiometric coefficients
Let's assign the standard enthalpy of formation of CO2(g) as ΔH°f(CO2).
The equation becomes:
-2923.0 kJ = 4ΔH°f(CO2) + 6ΔH°f(H2O) - 2ΔH°f(C2H6) - 7ΔH°f(O2)
However, before proceeding with the calculations, it is important to note that the standard enthalpy of formation of elements in their standard states is defined as zero. Therefore, the standard enthalpy of formation for O2(g) is zero. Additionally, let's assume the standard enthalpy of formation of H2O(g) is known (for example, -285.8 kJ/mol) as it was not given directly.
Plugging in these values, we have:
-2923.0 kJ = 4ΔH°f(CO2) + 6(-285.8 kJ/mol) - 2ΔH°f(C2H6)
Simplifying the equation further:
-2923.0 kJ = 4ΔH°f(CO2) - 1714.8 kJ/mol - 2ΔH°f(C2H6)
Rearranging the equation to solve for ΔH°f(CO2):
4ΔH°f(CO2) = -2923.0 kJ + 1714.8 kJ/mol + 2ΔH°f(C2H6)
4ΔH°f(CO2) = -1208.2 kJ/mol + 2ΔH°f(C2H6)
Thus, the standard enthalpy of formation of CO2(g) is equal to:
ΔH°f(CO2) = (-1208.2 kJ/mol + 2ΔH°f(C2H6))/4
Please provide the value for the standard enthalpy of formation of C2H6, and we can calculate the standard enthalpy of formation of CO2(g).
To determine the standard enthalpy of formation of CO2(g), we need to use the given information about the standard enthalpy change of the reaction and the standard enthalpies of formation for the other substances involved.
The standard enthalpy change, ΔH, for a chemical reaction can be calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. Mathematically, this can be expressed as:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
In this reaction, the standard enthalpy change (ΔH) is given as -2923.0 kJ. The standard enthalpies of formation (ΔHf) for the reactants and products involved in the reaction are as follows:
ΔHf(C2H6(g)) = -84.7 kJ/mol
ΔHf(O2(g)) = 0 kJ/mol
ΔHf(CO2(g)) = ?
ΔHf(H2O(g)) = -241.8 kJ/mol
Now, let's substitute these values into the equation and solve for the standard enthalpy of formation of CO2(g):
-2923.0 kJ = (4 * ΔHf(CO2)) + (6 * ΔHf(H2O)) - (2 * ΔHf(C2H6)) - (7 * ΔHf(O2))
-2923.0 kJ = (4 * ΔHf(CO2)) + (6 * (-241.8 kJ/mol)) - (2 * (-84.7 kJ/mol)) - (7 * 0 kJ/mol)
Simplifying the equation:
-2923.0 kJ = (4 * ΔHf(CO2)) - 1450.8 kJ + 169.4 kJ
Rearranging the equation:
(4 * ΔHf(CO2)) = -2923.0 kJ + 1450.8 kJ - 169.4 kJ
(4 * ΔHf(CO2)) = -1641.6 kJ
Dividing both sides of the equation by 4:
ΔHf(CO2) = -1641.6 kJ / 4
ΔHf(CO2) = -410.4 kJ
Therefore, the standard enthalpy of formation of CO2(g) is -410.4 kJ/mol.