Question

Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.
C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)

Answer 1

Answer:

$\Delta H_{rxn}=-2043.999kJ$

Explanation:

$\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]$

Where $n_{i}$ and $n_{j}$ are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).

$\Delta H_{f}^{0}$ is standard heat of formation and $\Delta H_{rxn}^{0}$ is standard enthalpy change for reaction at $25^{0}\textrm{C}$

So, $\Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

or, $\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]$

or, $\Delta H_{rxn}=-2043.999kJ$

Answer 2

The standard enthalpy change can be defined as the difference between the sum of product and the sum of the reactant -2043.999 kJ

The standard enthalpy Change for the reaction can be defined thus :

• $\Delta H_{rxn} = Σproduct - Σreactant$

Using the standard enthalpy values :

Σproduct = 3(-393.509) + 4(-241.818) = −2147.799

Σreactant = 1(-103.8) + 5(0) = -103.8

$\Delta H_{rxn} = −2147.799 - (-103.8)$

$\Delta H_{rxn} = −2147.799 + 103.8)$

$\Delta H_{rxn} = -2043.999 \: kj$

Therefore, the standard enthalpy change for the reaction is -2043.999 kj

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