Question #ae5d8

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Question #ae5d8

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Answer 1 · 2017-04-05T09:41:42

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Explanation:

First gather the information about the $DeltaHf$ of all the substances

$"substance" color(white)(mmmmm)DeltaHf$
$CO color (white)(mmmmmm)−110.5 kJ·mol^(−1)$
$O_2color (white)(mmmmmmmm) 0 kJ*mol^(-1)$
$CO_2 color (white)(mmmmm)−393.5kJ·mol^(−1)$

Always remember elements have $DeltaHf$ as 0

$DeltaH_"rxn" = DeltaH"f of product(s)" - DeltaH"f of reactant(s)"$

If there is a coefficient in front of the substance like C 2CO multiply the $DeltaHf$ by the coefficient

$DeltaH_"rxn" = −393.5kJ·mol^(−1) *2 -(−110.5 kJ·mol^(−1) + 0)2$

$DeltaH_"rxn" = −393.5kJ·mol^(−1)*2+110.5 kJ·mol^(−1)*2$

$DeltaH_"rxn" = -566kJ·mol^(−1)$

This means that the reaction is exothermic.

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Question #ae5d8

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