Calculate the enthalpy of the reactionSolution2B2O3s 6H2Og

Calculate the enthalpy of the reaction

2B2O3(s) +6H2O(g) -------------> 6O2(g) + 2B2H6 (g) delta H = 4070 kJ .............(1)

4B(s)+ 6H2(g)--------------->2B2H6 (g) delta H = 72 kJ ..............................(2)

subtracting equation (2) from equation (1) , we get

2B2O3(s) +6H2O(g) -------------> 6O2(g) + 4B(s) +6H2(g) delta H = 3998 kJ .............(3)

6H2(g) +3O2(g) -------------------> 6H2O(l) delta H = - 1710 kJ.............................(4)

6H2O(l) ---------------> 6H2O(g) delta H = 264 kJ...........................(5)

Adding equation (4) and (5) gives

6H2(g) +3O2(g) -------------------> 6H2O(g) delta H = -1446 kJ .................(6)

adding eqaution (3) and (6) , we get

2B2O3 (s) ----------------------> 3O2(g) + 4B(s) delta H = 2552kJ

hence for

3O2(g) + 4B(s) -------------------> 2B2O3 (s) delta H = -2552 kJ , which is the required heat of reaction.