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Heat of Reaction

Methane is a ...

Question

Methane is a commercial source of $H_{2}$ through the reaction
(I) $C H_{4} (g) + \frac{1}{2} O_{2} (g) \to C O (g) + 2 H_{2} (g)$
Based on the following thermochemical equations (II to IV)
(II) $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g); △ H = - 800 k J$

(III) $C H_{4} (g) + C O_{2} (g) \to 2 C O (g) + 2 H_{2} (g); △ H = + 235 k J$

(IV) $C H_{4} (g) + H_{2} O (g) \to C O (g) + 3 H_{2} (g); △ H = + 204 k J$
$△ H$ of equation (I) is

-35.75 kJ

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- 39.25 kJ

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- 25.75 kJ

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+ 25.75 kJ

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Solution

The correct option is B - 39.25 kJ
(II + III) gives V
$2 C H_{4} (g) + 2 O_{2} (g) \to 2 C O (g) + 2 H_{2} (g) + 2 H_{2} O (g);$
$△ H = - 800 + 235 = - 565 k J$
(V) $∴ C H_{4} (g) + O_{2} (g) \to C O (g) + H_{2} (g) + H_{2} O (g);$
$△ H = \frac{- 565}{2} k J = - 282.5 k J$
Eq. (IV) + Eq. (V) gives
$2 C H_{4} (g) + O_{2} (g) \to 2 C O (g) + 4 H_{2} (g);$
$△ H = (- 282.5 + 204) k J$
$∴ C H_{4} (g) + \frac{1}{2} O_{2} (g) \to C O (g) + 2 H_{2} (g);$
$△ H = - \frac{78.5}{2} = - 39.25 k J$

Hess’s Law of Constant Heat Summation:
Enthalpy change of a chemical equation is always constant whether the process occurs in a single step or several steps. In other words, the total enthalpy change in a reaction depends only upon the initial reactant and final product not upon the path.

Let say a reaction goes from A to D

$A \to D$ $Δ H$

$A \to B$ $Δ H_{1}$

$B \to C$ $Δ H_{2}$

$C \to D$ $Δ H_{3}$

According to the Hess’s Law of Constant Heat Summation

$Δ H = Δ H_{1} + Δ H_{2} + Δ H_{3}$

Chemical equations can be added or subtracted to yield the required equation.

Corresponding enthalpies are also manipulated in the same way to get the desired enthalpy.

Example:

$C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$ $Δ_{r} H$

Reaction 1:
$C (s) + O_{2} (g) \to C O_{2} (g)$ $Δ_{r} H_{1}$

Reaction 2:
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ $Δ_{r} H_{2}$

Now, if we reverse the reaction 2

Reaction 3:
$C O_{2} (g) \to C O (g) + \frac{1}{2} O_{2} (g)$ $- Δ_{r} H_{2}$

Now adding equation 1 and equation 3, we get

$C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$

So, $Δ_{r} H = Δ_{r} H_{1} - Δ_{r} H_{2}$

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Similar questions

H_{2}

C H_{4} (g) + \frac{1}{2} O_{2} (g) \to C O (g) + 2 H_{2} (g)

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g); △ H = - 800 k J

C H_{4} (g) + C O_{2} (g) \to 2 C O (g) + 2 H_{2} (g); △ H = + 235 k J

C H_{4} (g) + H_{2} O (g) \to C O (g) + 3 H_{2} (g); △ H = + 204 k J

△ H

(i) C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g); K_{1} (i i) C H_{4} (g) + H_{2} O (g) ⇌ C O (g) + 3 H_{2} (g); K_{2} (i i i) C H_{4} (g) + 2 H_{2} O (g) ⇌ C O_{2} (g) + 4 H_{2} (g); K_{3}

C O (g) + H_{2} O (g) ⇌ {C O}_{2} (g) + H_{2} (g); K_{1}

{C H}_{4} (g) + H_{2} O (g) ⇌ C O (g) + 3 H_{2} (g); K_{2}

{C H}_{4} (g) + 2 H_{2} O (g) ⇌ {C O}_{2} (g) + {4 H}_{2} (g); K_{3}

C O (g)

+ H_{2} O (g) ⇋ C O_{2} (g) + H_{2} (g); K_{1}

C H_{4} (g) + H_{2} O (g) ⇋ C O (g) + 3 H_{2} (g); K_{2}

C H_{4} (g) + 2 H_{2} O (g) ⇋ C O_{2} (g) + 4 H_{2} (g); K_{3}

C_{(g r a p h i t e)} + O_{2} (g) \to C O_{2} (g); Δ_{r} H^{o} = - 393.5 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ_{r} H^{o} = - 285.8 k J m o l^{- 1}

C O_{2} (g) + 2 H_{2} O (l) \to C H_{4} (g) + 2 O_{2} (g); Δ_{r} H^{o} = + 890.3 k J m o l^{- 1}

Δ_{r} H^{o}

298 K

C_{(g r a p h i t e)} + 2 H_{2} (g) \to C H_{4} (g)

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