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 don’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.
To calculate the enthalpy of formation (ΔHf) for the given reaction, you are on the right track. The enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements, with each element in its standard state.
You correctly set up the equation using the enthalpy of formation values for each compound:
ΔH = [1 * ΔH(Na2CO3) + 1 * ΔH(H2O(g)) + 1 * ΔH(CO2(g))] - [2 * ΔH(NaHCO3)]
However, for the next step, you need to substitute the correct values for the enthalpy of formation.
The enthalpy of formation for Na2CO3 is -1130.9 kJ/mol (as you correctly mentioned).
The enthalpy of formation for H2O (g) is -241.82 kJ/mol.
The enthalpy of formation for CO2 (g) is -393.5 kJ/mol.
But, the value you used for ΔH for NaHCO3 (-1766.22 kJ) seems to be incorrect. The enthalpy of formation for NaHCO3 is -950.2 kJ/mol.
Using these corrected values, your equation becomes:
ΔH = (1 * -1130.9 kJ) + (1 * -241.82 kJ) + (1 * -393.5 kJ) - (2 * -950.2 kJ)
Now, you can calculate the ΔH:
ΔH = -1130.9 kJ - 241.82 kJ - 393.5 kJ - (-1900.4 kJ)
= -1130.9 kJ - 241.82 kJ - 393.5 kJ + 1900.4 kJ
= 134.28 kJ
So, the enthalpy change for the reaction is 134.28 kJ.
Keep in mind that when calculating ΔHf, it is important to use the correct enthalpy of formation values for each compound involved in the reaction. Double-checking the values and ensuring their accuracy is key to obtaining the correct answer.