# Chemistry

posted by
**Laura** on
.

I am having trouble with this equation and calculating the enthalpy of formation.

2 NaHCO3 (s) --> Na2CO3 (s) + H2O (g) + CO2 (g) delta H129.2kj

I get as far as this step and then I donâ€™t know what to do next.

delta H= [1*delta H (Na2CO3) + 1* delta H(H2O (g)) + 1* delta H(CO2(g)) - [2* delta H (NaHCO3)]

= [ 1*( -1130.9kJ) + 1* ( -241.82kJ) + 1*( - 393.5kJ) ] = -1766.22kJ

Any help is greatly appreciated.