Ncert Exercises & Solutions

$∆_{f} U^{⊝}$ $∆_{f} U^{⊝}$ of formation of ${CH}_{4} (g)$ ${CH}_{4} (g)$ at certain temperature is $- 393$ $- 393$ $kJ$ $kJ$ ${mol}^{- 1}$ ${mol}^{- 1}$ . The value of $∆_{f} H^{⊝}$ $∆_{f} H^{⊝}$ is-

1. Zero

2. $< ∆_{f} U^{⊝}$ $< ∆_{f} U^{⊝}$

3. $> ∆_{f} U^{⊝}$ $> ∆_{f} U^{⊝}$

4. Equal to $∆_{f} U^{⊝}$ $∆_{f} U^{⊝}$

Hint:

∆_{f} H^{⊝} = ∆_{f} U^{⊝} + ∆ n_{g} RT

$∆_{f} H^{⊝} = ∆_{f} U^{⊝} + ∆ n_{g} RT$

The reaction is

The formula is as follows:

∆_{f} H^{⊝} = ∆_{f} U^{⊝} + ∆ n_{g} RT

$∆_{f} H^{⊝} = ∆_{f} U^{⊝} + ∆ n_{g} RT$

For the given reaction,

C (graphite)

$C (graphite)$

+

$+$

2 H_{2} (g) \to {CH}_{4} (g)

$2 H_{2} (g) \to {CH}_{4} (g)$

calculate the value of

∆ n_{g}

$∆ n_{g}$

∆ n_{g} = {(n_{p} - n_{r})}_{g} = 1 -2 = -1 ∆_{f} H^{⊝} = ∆_{f} U^{⊝} + ∆ n_{g} RT

$∆ n_{g} = {(n_{p} - n_{r})}_{g} = 1 -2 = -1 ∆_{f} H^{⊝} = ∆_{f} U^{⊝} + ∆ n_{g} RT$

∆ n_{g} = -1 ∆_{f} H^{⊝} = ∆_{f} U^{⊝} - 2 RT

$∆ n_{g} = -1 ∆_{f} H^{⊝} = ∆_{f} U^{⊝} - 2 RT$

∴

$∴$

∆_{f} H^{⊝} < ∆_{f} U^{⊝}

$∆_{f} H^{⊝} < ∆_{f} U^{⊝}$

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