We are given.... #Ca(s) + 1/2 O_2(g) -> CaO(s)# #;DeltaH_1^@ = -635.1*kJ*mol^-1# #CaCO_3(s) rarr CaO(s) + CO_2(g)# #;DeltaH_2^@ = 178.3 *kJ*mol^-1# What is the enthalpy of reaction for #Ca(s)+1/2O_2(g) +CO_2(g) rarr CaCO_3(s)#?

1 Answer
Jan 20, 2018

#DeltaH_"rxn"=-456.8*kJ*mol^-1#...

Explanation:

We want #DeltaH_"rxn"# for....

#Ca(s)+1/2O_2(g) +CO_2(g) rarr CaCO_3(s)#

We are given....

#Ca(s) + 1/2 O_2(g) -> CaO(s)# #;DeltaH_1^@ = -635.1*kJ*mol^-1#

#CaCO_3(s) rarr CaO(s) + CO_2(g)# #;DeltaH_2^@ = 178.3 *kJ*mol^-1#

And if take the former reaction, and add it the REVERSE of the latter reaction we get....

#Ca(s) + 1/2 O_2(g)+CaO(s) + CO_2(g)rarrCaCO_3(s)+CaO(s)#

And we cancel out common reagents....

#Ca(s) + 1/2 O_2(g)+cancel(CaO(s)) + CO_2(g)rarrCaCO_3(s)+cancel(CaO(s))#..

..to give....

#Ca(s) + 1/2 O_2(g) + CO_2(g)rarrCaCO_3(s)#..

And this is precisely the reaction whose enthalpy we wish to estimate...

And so we simply sum the GIVEN enthapies together and boserve the sign conventions .............

#DeltaH_"given reaction"^@=-635.1*kJ*mol^-1+178.3 *kJ*mol^-1=-456.8*kJ*mol^-1#