Enthalpy of formation of C 2 H 6 ( g ) at 300 K is − 20 kcal / mol then Δ U for formation of C 2 H 6 ( g ) will be ( R = 2 cal/mol-K)

Question

Enthalpy of formation of C 2 ​ H 6 ​ ( g ) at 300 K is − 20 kcal / mol then Δ U for formation of C 2 ​ H 6 ​ ( g ) will be ( R = 2 cal/mol-K)

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Question Explanation:

Calculating internal energy change from enthalpy of formation.

Concept:

Relationship between Enthalpy ( $Δ H$ ) and Internal Energy ( $Δ U$ ).

Solution:

The enthalpy change and internal energy change for a gaseous reaction are related by the formula involving the change in the number of moles of gaseous components. We first write the balanced chemical equation for the formation of ethane ( $C_{2} H_{6}$ ) from its constituent elements in their standard states to find the value of $Δ n_{g}$ .

Reaction: $2 C (S) + 3 H_{2} (g) \to C_{2} H_{6} (g)$

Change in gaseous moles ( $Δ n_{g}$ ): $n_{p ro d u c t} - n_{re a c t an t} = 1 - 3 = - 2$

Formula: $Δ H = Δ U + Δ n_{g} RT$

Given: $Δ H = - 20 Kcal/mol = - 20000 cal/mol$ ; $R = 2 cal/mol \cdot K$ ; $T = 300 K$

$- 20000 = Δ U + (- 2 \times 2 \times 300)$

$- 20000 = Δ U - 1200$

$Δ U = - 20000 + 1200 = - 18800 cal/mol$

$Δ U = - 18.8 kcal/mol$

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