Below is an example of how to use Hess\' Law to determine the enthalpy of a reac

Question

Below is an example of how to use Hess' Law to determine the enthalpy of a reaction. Fill-in the blanks with the correct energies in kJ CH_4 (g) + 2 O_2_(g) rightarrow CO_2_(g) + 2 H_2 O_(l) delta H degree = ? Use the following information to calculate delta H degree in KJ for this reaction. CH_4(g) + O_2(g) rightarrow CH_2 O(g) + H_2 O(g) delta H degree = -275.6 KJ CH_2 O_(g) + O_2_(g) rightarrow CO_2 (g) + h_2 O(g) delta H degree = -526.7 kJ H_2 O(l) rightarrow H_2 O(g) delta H degree = 44.0 kJ SOLUTION CH_4(g) + O_2(g) rightarrow CH_2 O(g) + H_2 O(g) delta H degree = CH_2 O(g) + O_2(g) rightarrow CO_2(g) + H_2 O(g) delta H degree = 2(H_2 O(g) rightarrow H_2 O(l) delta H degree = CH_4(g) + 2 O_2(g) rightarrow CO_2(g) + 2 H_2 O(l) delta H degree =

Explanation / Answer

From the example

the first reaciton is left unmodified, so H1 = -275.6 kJ remains

The second reaciton is also left alone so, H2 = -526.7

finally, there is water shift, which is INVERTED and multiplied by 2, then H3 = -44*2 = -88

The overall reaction is laready written

the Hvalue is given by

Hrxn = H1+H2+H3 = 275.6 -526.7 - 88 = -890.3 kJ

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