Question
Calculate the standard enthalpy change for the reaction 2C8H18(l)+17O2(g)> 16CO(g) + 18H2O(l)
Given
2C8H18(l)+25O2(g)>16CO2(g)+18H2O(l) deltaH=-11020 kj/mol
2CO(g)+ O2(g)>2CO2(g)
Delta H=-566.0 kj/mol
The > is supposed to be an arrow :)
Given
2C8H18(l)+25O2(g)>16CO2(g)+18H2O(l) deltaH=-11020 kj/mol
2CO(g)+ O2(g)>2CO2(g)
Delta H=-566.0 kj/mol
The > is supposed to be an arrow :)
Answers
delta Hrxn = (n*DHf products) - (n*DHf reactants)
641.7 KJ
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