calculate the enthalpy of formation of methane from the following data:

C(s)+O2(g)=CO2(g)                       deltaH=-393.5 kJ

2H2(g)+O2(g)=2H2O(l)                  deltaH=-571.8kJ

CH4(g)=2O2(g)=CO2(g)=2H2O(l)          deltaH=-890.3

We have to calculate the enthalpy for formation of methane that is the following reaction :

 C(s) + 2H2 (g) → CH4 (g)  (a)

In this question we will manipulate the data that is provided to for different equations and get the result. We are given that

 C (s) + O2(g)  → CO2(g)  ΔH=-393.5 kJ  (1)

 2H2(g) + O2(g)  → 2H2O(l)  ΔH=-571.8kJ  (2)

 CH4(g)+ 2O2(g)  → CO2(g) + 2H2O(l)  ΔH=-890.3 kJ    (3)

Let us add equation (1) and (2). The corresponding ΔH values for these two reactions would also be added. So we get

 C (s) + 2H2(g) + 2O2(g)  → CO2(g) + 2H2O(l)    ΔH = -965.3 kJ

Now if we subtract equation (3) from the above equation, we will get the desired equation that is equation (a). Hence the corresponding ΔH values would also be subtracted. Thus we get  

 C(s) + 2H2 (g) → CH4 (g)  ΔH = -175.0 kJ

So the enthalpy of formation of methane is -175 kJ. 

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