Question

Calculate the heat of combustion of glucose from the following data-
C (graphite) + O2(g) → CO2(g) ∆H = -395.0 KJ
H2(g) +02(g) → H2O(1) ∆H = -269.4 KJ
6 C(graphite) + 6H2(g) + 302(g) → C6H1206(s) ∆H = -1169.8 KJ​

Answer 1

Answer:

For calculating the Enthalpy of Combustion of Glucose i.e. ΔH for the following reaction

C6H12O6 + 6O2 → 6CO2 + 6H2O  (a)

We will calculate the value of ΔH for the above reaction by adding and subtracting the provided equations. The corresponding ΔH values for the reactions will also be multiplied, added and subtracted in the same manner. We are given that

1. C(graphite) + O2(g) ------------> CO2(g)  ΔH0 = -395 kJ

2. H2(g) + 1/2O2(g) -----------> H2O(l)  ΔH0 = -286 kJ

3. 6C(graphite) + 6H2(g) +3O2(g) ----------> C6H12O6(s)  ΔH0 = -1170 kJ

To obtain the required equation that is (a), we will adopt the following steps

(i) multiplying equation (1) and (2) with 6 and adding the equation. The corresponding ΔH values would also be multiplied by the respective numbers and added. Thus the net equation and the corresponding ΔH that we will obtain are as follows :

6C (graphite) + 6H2(g) + 9O2 (g) → 6CO2 + 6H2O  ΔH0 = -4080 kJ  (b)

(ii) Now, equation (3) will be subtracted from the above equation to obtain equation (a), which is the required equation. Therefore the respective ΔH0 would also be subtracted. Hence we will obtain

C6H12O6 + 6O2 → 6CO2 + 6H2O  ΔH0 = -2910 kJ  

Therefore, the heat of combustion of the above reaction is -2910 kJ