Ncert Exercises & Solutions

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

1. Zero

2. $< ∆_{f} U^{⊝}$

3. $> ∆_{f} U^{⊝}$

4. Equal to $∆_{f} U^{⊝}$

Hint:

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

The reaction is

The formula is as follows:

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

For the given reaction,

C (graphite)

+

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

calculate the value of

∆ n_{g}

∆ 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

∴

∆_{f} H^{⊝} < ∆_{f} U^{⊝}