Hess's Law Help

Started by RAMPHARMD
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RAMPHARMD

UHCOP Pharm.D. Student
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Hey guys, I'm working on a lab report and I'm stuck. I have spent over an hour trying to figure this stinker out and still no luck. See if you guys can help out.

Find ∆H for the reaction, HCl(g) + H2O(l) à HNO2(l) + Na2O (s), given:

2 NaCl(s) + H2O(l) à 2 HCl(g) + Na2O(s); ∆H = 507 kJ
NO(g) + NO2(g) + Na2O(s) à 2 NaNO2(s); ∆H = -427 kJ
NO(g) + NO2(g) à N2O(g) + O2(g); ∆H = -43 kJ
2 HNO2(l) à NO2(g) + O2(g) + H2O(l); ∆H = 34 kJ

Thanks in advance!
 
play around with it to cancel out some elements. for example to cancel out elements you need to multiply second row by minus 1 (-1) and so it becomes 2 NaNO2(s)àNO(g) + NO2(g) + Na2O(s) ??, so enthalpy will be +427. multiply the first equation by 2 or 1/2??? and 507*2 kJ or (1/2kj) so coefficients can cancel out as well. play with it until you end up with equation that was given to you at the begining; add all the enthalpy and you will get your final answer. I don't have paper otherwise i would have solved it for you.
Hey guys, I'm working on a lab report and I'm stuck. I have spent over an hour trying to figure this stinker out and still no luck. See if you guys can help out.

Find ∆H for the reaction, HCl(g) + H2O(l) à HNO2(l) + Na2O (s), given:

2 NaCl(s) + H2O(l) à 2 HCl(g) + Na2O(s); ∆H = 507 kJ
NO(g) + NO2(g) + Na2O(s) à 2 NaNO2(s); ∆H = -427 kJ
NO(g) + NO2(g) à N2O(g) + O2(g); ∆H = -43 kJ
2 HNO2(l) à NO2(g) + O2(g) + H2O(l); ∆H = 34 kJ

Thanks in advance!