**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.

CH_{4} (g) + 2O_{2} —> CO_{2} (g) + 2H_{2}O (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.

H_{2}O (s) —> H_{2}O (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.

H_{2}O (s) —> H_{2}O (l) ; ΔH= +6.01 kJ

H_{2}O (l) —> H_{2}O (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.

H_{2}O (s) —> H_{2}O (l) ; ΔH= +6.01 kJ

H_{2}O (l) —> H_{2}O (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…

H_{2}O (s) —> H_{2}O (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.

2H_{2}O (s) —> 2H_{2}O (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 H_{2}O (s) —> ½ H_{2}O (l) ; ΔH= +3.005kJ

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

2H_{2}O (l) —> 2H_{2}O (s) ; ΔH= – 12.02 kJ Notice the – sign!