How to write a the thermochemical equation
Define thermochemical equation:
Thermochemical equation is the chemical equations for a reaction in which equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation.
2Na(s) + 2H2O (l) ..................> 2NaOH (aq) + H2 (g) ΔH = -367.5 kj
The information that we can draw from this equation is that 2 mol of sodium reacts with 2 mol of water to produce 2 mol of sodium hydroxide and 1 mol of hydrogen gas, and 367.5 kj of heat evolves.
Lets see a solved problem:A
n aqueous sodium hydrogen carbonate reacts with hydrochloric acid to produce aqueous sodium chloride,water and carbon dioxide gas.The reaction absorbs 11.8 kJ of heat at constant pressure for each mole of sodium hydrogen carbonate.Write the thermochemical equation for the reaction.
Step1 : First write the balanced chemical equation
NaHCO3 (aq) + HCl (aq) ------------------> NaCl (aq) + H2O(l) + CO2 (g)
Note: if you don't know how to balance equation and writing the formula and name of compounds look at our tutorials
Step 2 : As you can see in the problem the reaction absorbs heat, the enthalpy of reaction for molar amounts of this equation is +11.8 kJ( as reaction absrobs heat there is a + sign), so final thermochemical equation is:
NaHCO3 (aq) + HCl (aq) ------------------> NaCl (aq) + H2O(l) + CO2 (g) ΔH = + 11.8 kJ
Rule for manipluation thermochemical equations :
There are following two rules for manipulation thermochemical equations:
1. When a thermochemical equation is multiplied by any factor, the value of ΔH for the new equation is obtained by multiplying the value of ΔH in the original equation by that same factor.
2. When a chemical equation is reverse,the value of ΔH is reversed in sign.
Lets see an example :
Problem : When 2 mol H2(g) and 1 mol of O2(g) react to give liquid water , 572 kJ of heat evolves,
2H2(g) O2(g) ...................> 2 H2O(l) ΔH = -572 kJ
write this equation for 1 mol of liquid water.Give the reverse equation in which 1 mol of liquid water dissociates into hydrogen and oxygen.
Step1: you multiply the coefficents and ΔH by 1/2
1/2 ( 2H2(g) + O2(g) ...................> 2 H2O(l) ΔH = -572 kJ)
H2(g) + 1/2 O2(g) ...................> H2O(l) ΔH = -286 kJ
notice tht ΔH value is also half of its previous value.
Step2: Reversing the equation you get
H2O(l) .......................... > H2(g) + 1/2O2(g) ΔH = + 286 kJ
notice that sign of ΔH changes when we have reversed the equation.