Writing Thermochemical Equations

Rules for Writing Thermochemical Equations: 

(1)   When heat is lost, the ΔH value is negative. The heat value may also be given in the equation itself as a product.

 CH₄ (g) + 2O₂ —> CO₂ (g) + 2H₂O (l)            ΔH= -890.4 kJ

 (2)   When heat is gained, the ΔH value is positive. The heat value may also be given in the equation itself as a reactant.

H₂O (s) —> H₂O (l)          ΔH= +6.01 kJ

(3)  Since heat is a state function, the ΔH value for the same equation may be different if it occurs in different physical states. Therefore, all physical states must be written in the equation.

H₂O (s) —> H₂O (l) ; ΔH= +6.01 kJ

H₂O (l) —>  H₂O (g) ; ΔH= +44 kJ

 The same reactant and product of water but in different physical states yields different enthalpy or ΔH values!

(4) If a reaction is reversed then the enthalpy (ΔH) value will also be reversed. Hence a + becomes a – and vise versa.

 H₂O (s) —> H₂O (l) ; ΔH= +6.01 kJ

 H₂O (l) —>  H₂O (s) ; ΔH= – 6.01 kJ

 (5) If we change the stoichiometric coefficients in the chemical reaction, then we also change the enthalpy (ΔH) value proportionally! Therefore, if you double the reactants, you will double the products and also double the enthalpy (ΔH) and so on…

  H₂O (s) —> H₂O (l) ; ΔH= +6.01 kJ

 a.) If we double the coefficients, we must also double the ΔH. Think of it as multiplying the entire equation by 2.

 2H₂O (s) —>  2H₂O (l) ; ΔH= +12.02 kJ

 b.) If we triple the coefficients, we must also triple the ΔH. Think of it as multiplying the entire equation by 3.

 3H2O (s) —>  3H2O (l) ; ΔH= +18.03 kJ

c.) If we half the coefficients, we must also half the ΔH. Think of it as multiplying the entire equation by ½.

½ H₂O (s) —> ½ H₂O (l) ; ΔH= +3.005kJ

 d.) If we double the coefficients AND reverse, we must also double the ΔH AND REVERSE the sign of ΔH.

  2H₂O (l) —> 2H₂O (s) ; ΔH= – 12.02 kJ      Notice the – sign!