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Methane (CH4) can burn in oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) as outlined in the following equation: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O.Calculate the standard enthalpy change associated with this reaction at 25 °C using thefollowing enthalpies of formation:CH4 ∆Hf = -62.7 kJ·mol–1, CO2 ∆Hf = -393.5 kJ·mol–1, H2O ∆Hf = -285.8 kJ·mol–1.

Sagot :

Answer:

-902.4kJ/mol

Explanations:

Give the chemical reaction below as shown;

[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex]

The standard enthalpy change is expressed as;

[tex]\begin{gathered} \triangle H^o=\triangle H_{f(product)}^o-\triangle H^0_{f(reactant)} \\ \triangle H^o=[\triangle H_{CO2}^o+2\triangle H_{H2O}^o]-[\triangle H_{CH4}^o+2\triangle H_{O2}^o] \end{gathered}[/tex]

Substitute the given enthalpy change

[tex]\begin{gathered} \triangle H^o=[-393.5+2(-285.8)]-[-62.7+2(0)] \\ \triangle H^o=[-393.5-571.6]+62.7 \\ \triangle H^o=-965.1+62.7 \\ \triangle H^o=-902.4kJmol^{-1} \end{gathered}[/tex]

Therefore the standard enthalpy change associated with this reaction is -902.4kJ/mol

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