ch4(g)+ 2 o2(g) + co2(g) + 2 h20 (l)
calculate the ∆h° for the reaction round your answer to the correct number of decimal places based on the table values.

∆H° reaction = -890.3 kJ

Explanation:

The given equation is :

Now ,

O2 is in the standard state so its  ∆H° is zero.

∆H° is calculated by considering the formation of CO2 , H2O and CH4 .

..........∆H°a = -393.5 kJ

.....∆H°b = -285.8 kJ

..........∆H°c = -74.8 kJ

Multiply equation of water H2O by 2

and reverse the direction of equation of CH4

Hence the sign of ∆H°c = +74.8 kJ becomes +ve.

We are doing this because CH4 is to be in the reactant side not  in the product side.

∆H° reaction = ∆H°a +2(∆H°b) -∆H°c

∆H° reaction = -393.5 - 2(285.8) + 74.8

∆H° reaction = -890.3 kJ