Ncert Exercises & Solutions

Based on the reactions, ascertain the valid algebraic relationship:
$(i)$ $C (g)$ $+ 4 H$ $(g)$ $\to C H_{4} (g);$
$∆ H_{r}$ $= x$ $k J$ $m o l^{- 1}$
$(i i)$ $C (g r a p h i t e)$ $+$ $2 H_{2}$ $(g)$ $\to$ $C H_{4};$
$∆ H_{r}$ $=$ $y$ $k J$ $m o l^{- 1}$

1.	x = y	2.	x = 2y
3.	x > y	4.	x < y

Hint: Enthalpy of formation of the compound can be positive or negative
depending on the type of reaction.

Explanation:

In a chemical reaction, the reactant first converts into its atomic state,
then combined, and formed a product.

In the first reaction, C and H are present in their atomic state hence
they will combine and released X amount of energy.

In the second reaction, $H_{2}$ molecule first converts into atomic hydrogen and Graphite first converts into Carbon(g) then combined with a carbon atom and formed a product.

The overall $∆ H_{r}$ for both reactions are as follows:

${i. \Delta H_r=x=- energy \ of \ the \ bonds \ being \ formed \ (four \ \mathrm{C}-\mathrm{H} \ bond \ formed)}$
$ii. \Delta \mathrm{H}_{\mathrm{r}}=\mathrm{y}= \ energy \ of \ the \ bond \ being \ broken \ ( \mathrm{H}-\mathrm{H} \ bond \ break)+\ energy~ for ~convert~graphite~ to ~C(g) \ - $
$ \ energy \ of \ the \ bonds \ being \ formed \ (four \ \mathrm{C}-\mathrm{H} \ bond \ formed) $

In both the reaction's energy of the bonds being formed is equal
but in the second reaction, extra energy is used to break the hydrogen molecule bond. Thus, x>y.

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