Calculate the enthalpy (ΔHo) of reaction (kJ) for the equation as written. If the answer is negative, enter the sign and then the magnitude.
2 HCOOH(l) + O2(g) → 2 CO2(g) + 2 H2O(l) .
ΔHof (kJ/mol)
HCOOH(l) -409
O2(g) 0
CO2(g) -393.5
H2O(l) -285.83
the answer is 501.7...how do you come to this??
To calculate the enthalpy change (ΔHo) of the reaction, you need to consider the enthalpy of formation (ΔHof) for each reactant and product.
Given that:
ΔHof (kJ/mol)
HCOOH(l) = -409
O2(g) = 0
CO2(g) = -393.5
H2O(l) = -285.83
We can use the equation:
ΔHo = Σ(nΔHof(products)) - Σ(nΔHof(reactants))
Plugging in the values, we get:
ΔHo = (2 * -393.5) + (2 * -285.83) - (2 * -409) - 0
ΔHo = -787 - 571.66 + 818
ΔHo = 501.34 kJ/mol
The enthalpy change of the reaction is +501.34 kJ. Since the answer should be negative according to the given equation, the sign should be negative, and the magnitude should be 501.34 kJ. Therefore, the final answer should be -501.34 kJ.