CHEM 1040 Lecture Notes - Lecture 11: Joule

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11 Oct 2018
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Thermo - part ii solutions to assigned problems. Given: (a) n2(g) + o2(g) no(g); h rxn(a) = +90. 4 kj (b) n2(g) + o2(g) no2(g); h rxn(b) = +33. 2 kj. Calculate hrxn for: 2no(g) + o2(g) 2no2(g) Steps: for equation (a), no appears on the opposite side of the equation relative to the equation we want and there is only 1 mole. So we multiply (b) by 2 to get equation (d), i. e. , (d) = 2 (b): N2(g) + 2o2(g) 2no2(g); h rxn(d) = 2 (+33. 2 kj) = 66. 4 kj: now add (c) and (d) together: Question: calculate the heat required to decompose limestone: caco3(s) cao(s) + co2(g) H f [co2(g)] = 393. 5 kj/mol: the long way (hess"s law): Ca(s) + c(s) + 3/2o2(g) caco3(s); (1) (2) (3) Therefore, flip equation (1) and add (2) and (3): Caco3(s) ca(s) + c(s) + 3/2o2(g) ; Caco3(s) cao(s) + co2(g) : the short way:

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