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

$∆ n_{g}$

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

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

$∴$

$∆_{f} H^{⊝} < ∆_{f} U^{⊝}$