Respuesta :

shinmin

The gas that experiences combustion yields 2428 J of heat, which suggests that q will be negative.


q=−2428 J


Similarly, work done by the system means that work was done by the gas on the surrounds, which once again suggests a minus sign.


w=−6 kJ


Since the gas burns at constant pressure, the heat given off will also be the enthalpy change, ΔH


ΔH=q → at constant pressure;


Change J to kJ to get


2428J x 10^−3kJ/ 1J = 2.428 kJ


Since you're distributing with 1 mole, you can write:


ΔH=−2.428 kJ/mol

Now use this equation to determine ΔE

ΔE=q+w

ΔE=−2.428 kJ+(−6 kJ)=−8.428 kJ

W0lf93
The gas that undergoes combustion produces 2428 J of heat, which implies that  
q will be negative.  
q =â’ 2428 J  
Likewise, work done by the system means that work was performed by the gas on the surroundings, which once again implies a minus sign. 
 w =â’6 kJ 
 Because the gas burns at constant pressure, the heat given off will also be the enthalpy change, Δ H 
 Î”H = q → at constant pressure; 
 Convert J to kJ to get 
 2428Jâ‹…10-3k/1J =2.428 kJ 
 Since you're dealing with 1 mole, you can write 
 Î” H = â’ 2.428 kJ/mol 
 Now use this equation to determine Δ E 
 Î” E = q + w 
 Î” E = â’ 2.428 kJ + (â’6 kJ) = â’ 8.428 kJ  
SIDE NOTE I'll leave the rounding to the correct number of sig figs to you

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