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Zinaida [17]
3 years ago
13

Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this lis

t of thermodynamic properties.
C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)
Chemistry
1 answer:
weqwewe [10]3 years ago
4 0

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

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