Consider the reactions given below. on the basis of these reactions find out which of the algebric relation given in options (i) to (iv) is correct ? (i) $C (g) + 4 H (g) \to C H_{4} (g); Δ H = x k J$ $m o l^{- 1}$ (ii) $C (g r a p h i t e) + 2 H_{2} (g) \to C H_{4} (g); Δ H = y k J$ $m o l^{- 1}$

x = y

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x = 2 y

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x > y

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x < y

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Solution

The correct option is A $x > y$
In both the reactions the product formed is same but in reaction (i), both the reactants are present in their nascent form and energy is not required for changing the state of reactants or breaking of bond.
While in reaction (ii), carbon is present in solid state which has to be converted into gaseous state. Also the $H - H$ bond has to be broken. Both the processes involve consumption of energy.
Hence, the enthalpy of reaction (i) is greater than the enthalpy of reaction (ii). So, $x > y$ .

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Similar questions

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

C_{(g r a p h i t e, s)} + 2 H_{2 (g)} \to C H_{4 (g)}; Δ_{r} H = y k J m o l^{- 1}

(i) C_{(g)} + 4 H_{(g)} \to C H_{4 (g)}; △_{r} H = x k J m o l^{- 1} (i i) C (g r a p h i t e, s) + 2 H_{2 (g)} \to C H_{4 (g)}; △_{r} H = y k J m o l^{- 1}

C (s) + 2 H_{2} (g) ⟶ C H_{4} (g); Δ H = - X k c a l

C (g) + 4 H (g) ⟶ C H_{4} (g); Δ H = - X_{1} k c a l

C H_{4} (g) ⟶ C H_{3} (g) + H (g); Δ H = + Y k c a l

C - H

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

C (g r a p h i t e) + \frac{1}{2} O_{2} (g) \to C O (g), Δ H = y k J m o l^{- 1}

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ H = z k J m o l^{- 1}

C (g) + 4 H (g) ⟶ C H_{4} (g)

Δ H

Δ U

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