Question

CH4(g)+ 2 O2(g) + CO2(g) + 2 H20 (l)

Answer 1

Answer:

∆H° reaction = -890.3 kJ

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

$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O(l)$.....∆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