Calculate the delta H for 2CO + 2NO = 2CO2 + N2

Question

Given the following data:

2CO + O2 = 2CO2 delta H= -566.0kJ

N2 + O2 = 2NO delta H= 180.6kJ

Answer 1

Rearrange each individual chemical equation so that in the end, when you add up the equations, you end up with the desired equation: 2CO + 2NO --> 2CO2 + N2

Equation 1: 2CO + O2 --> 2CO2

Equation 2: N2 + O2 --> 2NO

In the final equation, there is no O2, so you flip the Equation 2 to cancel out the O2.

2CO + O2 --> 2CO2

2NO --> N2 + O2

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2CO + 2NO --> 2CO2 + N2

Whenever you flip an equation, you flip the sign of its enthalpy value. So, delta H for Equation 2 would change from 180.6 to -180.6 kJ

Then, just add up the new enthalpy values: -566 + (-180) = -746.6 kJ

Answer 2

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