given C12H22O11+H2O----C6H12O6+C6H12O6 heat combustion0f 2826.7kj/mol what is the standard enthalpy
To find the standard enthalpy (ΔH) of the given reaction, we need to use the standard enthalpies of formation for the compounds involved. The standard enthalpy of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
The balanced equation for the combustion reaction is:
C12H22O11 + H2O → C6H12O6 + C6H12O6
The standard enthalpy change (ΔH) for the reaction can be calculated using the following equation:
ΔH = Σ(nΔHf(products)) - Σ(mΔHf(reactants))
Where:
ΔH = Standard enthalpy change
n = Stoichiometric coefficient for the products
m = Stoichiometric coefficient for the reactants
ΔHf = Standard enthalpy of formation
Looking up the standard enthalpies of formation for the compounds involved:
ΔHf(C12H22O11) = -915 kJ/mol
ΔHf(H2O) = -286 kJ/mol
ΔHf(C6H12O6) = -1274 kJ/mol
Now, plugging the values into the equation:
ΔH = (2 × ΔHf(C6H12O6)) - (ΔHf(C12H22O11) + ΔHf(H2O))
ΔH = (2 × -1274 kJ/mol) - (-915 kJ/mol + -286 kJ/mol)
= -2548 kJ/mol + 1201 kJ/mol + 286 kJ/mol
= -1061 kJ/mol
Therefore, the standard enthalpy change (ΔH) for the combustion reaction is -1061 kJ/mol.
To determine the standard enthalpy change (∆H°) for the given reaction, you need to use the standard enthalpy of formation (∆H°f) values for the reactants and products.
1. Start by writing the balanced equation for the reaction:
C12H22O11 + H2O → C6H12O6 + C6H12O6
2. Determine the standard enthalpy of formation values for each compound. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states at standard conditions (298K and 1 atm).
The standard enthalpy of formation values for the compounds involved in the equation are:
∆H°f[C12H22O11] = -2220 kJ/mol
∆H°f[H2O] = -285.8 kJ/mol
∆H°f[C6H12O6] = -1274.9 kJ/mol
3. Calculate the ∆H° for the reaction using the ∆H°f values. The enthalpy change for the reaction is equal to the sum of the ∆H°f values for the products minus the sum of the ∆H°f values for the reactants:
∆H° = ∑(∆H°f products) - ∑(∆H°f reactants)
∆H° = 2(-1274.9 kJ/mol) - (-2220 kJ/mol) - (-285.8 kJ/mol)
∆H° = -2549.8 kJ/mol + 2220 kJ/mol + 285.8 kJ/mol
∆H° = -43.2 kJ/mol
Therefore, the standard enthalpy change for the combustion of 1 mole of C12H22O11 is -43.2 kJ/mol. Note that the negative sign indicates an exothermic reaction.