Question

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.
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Answer 1

∆H° reaction = -890.3 kJ

Step-by-step explanation:

The given equation is :

`CH_(4)(g)+2O_(2)(g)\rightarrow CO_(2)(g)+2H_(2)O(l)`

Now ,

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

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

`C(s)+H_(2)(g)\rightarrow CO_(2)`..........∆H°a = -393.5 kJ

`H_(2)+(1)/(2)O_(2)\rightarrow H_(2)O(l)`.....∆H°b = -285.8 kJ

`C+2H_(2)\rightarrow CH_(4)(g)`..........∆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