At constant pressure and 25 degrees C, what is delta H for the reaction
2C2H6(g) + 7O2(g) --yields-- 4CO2(g) + H2O(l)
if the complete consumption of 11.3 g of C2H6 releases -586.3 kJ of heat energy?
(11.3=.38 mol C2H6)
-441 kJ
I get -3113 as my answer when I plug in 30.6 as the molar mass when the teacher said it should be around -225 as the final answer?
To calculate the enthalpy change (ΔH) for the given reaction, you need to use the concept of stoichiometry and the known energy change.
First, determine the number of moles of C2H6 consumed in the reaction. You have already calculated that 11.3 g of C2H6 corresponds to 0.38 mol.
Next, you can use the stoichiometric coefficients from the balanced chemical equation to determine the moles of CO2 and H2O produced. According to the equation, 2 mol of C2H6 produces 4 mol of CO2 and 1 mol of H2O. Therefore, 0.38 mol of C2H6 will produce (0.38 mol)(4 mol CO2 / 2 mol C2H6) = 0.76 mol of CO2 and (0.38 mol)(1 mol H2O / 2 mol C2H6) = 0.19 mol of H2O.
Now that you have the moles of CO2 and H2O produced, you can use the energy change (-586.3 kJ) to calculate the enthalpy change per mole of the reaction. Since 0.38 mol of C2H6 produces 0.76 mol of CO2, the enthalpy change per mole of the reaction is (-586.3 kJ) / (0.38 mol) = -1540.8 kJ/mol.
Therefore, at constant pressure and 25°C, the ΔH for the reaction 2C2H6(g) + 7O2(g) → 4CO2(g) + H2O(l) is -1540.8 kJ/mol.