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Data: A H f values: CH 4( g), -74.8 kJ; CO 21 g), -393.5 kJ; H 20( 1), -285.8 kJ. Using the A H f data above, calculate A H xn for the reaction below. Reaction: CH 4( 9) + 20 2( 9) => CO 2(g) + 2H 2011) Selected Answer: d. -890.3 kJ Correct Answer: d. -890.3 kJ

Respuesta :

Answer:

[tex]\Delta H_{rxn}[/tex] for the given reaction is -890.3 kJ

Explanation:

[tex]\Delta H_{rxn}=\sum n_{i}.\Delta H_{f}(product)_{i}-\sum n_{j}.\Delta H_{f}(reactant)_{j}[/tex]

where [tex]n_{i}[/tex] and [tex]n_{j}[/tex] represents number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence [tex]\Delta H_{rxn}=[1mol\times \Delta H_{f}(CO_{2})_{g}]+[2mol\times \Delta H_{f}(H_{2}O)_{l}]-[1mol\times \Delta H_{f}(CH_{4})_{g}]-[2mol\times \Delta H_{f}(O_{2})_{g}][/tex]

so, [tex]\Delta H_{rxn}=[1mol\times -393.5kJ/mol]+[2mol\times -285.8kJ/mol]-[1mol\times -74.8kJ/mol]+[2mol\times 0kJ/mol]=-890.3 kJ[/tex]

So, Correct answer is -890.3 kJ

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