when calculating the standard enthalpy of reaction I got
∆H = 4(91.3 kJ/mol of NO) + 2(-114.1 kJ) + (-110.2 kJ) = 26.8 kJ
26.8 kJ / 2 mol N2O5 = 13.4 kJ / mol
I was wondering if someone could help me check my work and explain how to get 11. 4 kJ/mol
Textbook 4D.23
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Simone Byun 1F
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Re: Textbook 4D.23
The enthalpy of formation for NO is 90.25 kJ, not 91.3. Your calculations would instead be:
∆H = 4(90.25 kJ/mol of NO) + 2(-114.1 kJ) + (-110.2 kJ) = 22.6 kJ
22.6 kJ / 2 mol N2O5 = 11.3 kJ / mol
∆H = 4(90.25 kJ/mol of NO) + 2(-114.1 kJ) + (-110.2 kJ) = 22.6 kJ
22.6 kJ / 2 mol N2O5 = 11.3 kJ / mol
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