Using the table below, what is the change in enthalpy for the following reaction when all reactants and products are in the gaseous state?
Nitrogen Dioxide Nitrogen Monoxide + Oxygen

Substance ΔHf (kJ/mol)
NO2 (g) 33.85
NO (g) 90.37
O2 (g) 0.0
O3 (g) 142.0

Question

Using the table below, what is the change in enthalpy for the following reaction when all reactants and products are in the gaseous state?
Nitrogen Dioxide  Nitrogen Monoxide + Oxygen


Substance	ΔHf (kJ/mol)
NO2 (g)	        33.85
NO (g)	      90.37
O2 (g)	       0.0
O3 (g)	       142.0

jfraser · Answer

enthalpy of rxn is always$$\Delta H_{rxn} = \Sigma \Delta H_{products} - \Sigma \Delta H_{reactants}$$
add up the DH of all the products, add up all the DH of the reactants, and subtract the 2

anonymous · Answer

255.0 kJ

anonymous · Answer

??

anonymous · Answer

these are the choices:

		
113.0 kJ
 		
255.0 kJ
 		
56.5 kJ
 		
−113.0 kJ

anonymous · Answer

Is my answer correct or did i do something wrong?

jfraser · Answer

$$2NO_2(g) \rightarrow 2NO(g) + O_2(g)$$

$$\Delta H = \Sigma [(\Delta H O_2) + 2(\Delta H NO)] - \Sigma [2(\Delta HNO_2)]$$ I get 113 kJ

anonymous · Answer

113 is the answer