Calculate the heat of formation of CO2 by following data- - Hc C2H6 - -397-8 Kcal - Hf C2H6 - -21-1 Kcal - Hf H2O - -94-1 Kcal

Given: Combustion of C_2H_6 C_2H_6_( g ) + 72O_2_( g )⟶ 2 CO_2_( g ) + 3 H_2O_( l ) ΔH = 397.8 kcal ..... ( 1 ) 2 C_( s ) + 3 H_2_( g )⟶C_2H ...

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Byju's Answer

Standard XII

Physics

Equipartition Theorem

Calculate the...

Question

Calculate the heat of formation of $C O_{2}$ by following data.
( $Δ$ Hc) $C_{2} H_{6}$ = $- 397.8$ Kcal
( $Δ$ Hf) $C_{2} H_{6}$ = $- 21.1$ Kcal
( $Δ$ Hf) $H_{2} O$ = $- 94.1$ Kcal

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Solution

Given:-
Combustion of $C_{2} H_{6}$ -
${C_{2} H_{6}}_{(g)} + \frac{7}{2} {O_{2}}_{(g)} ⟶ 2 {C O_{2}}_{(g)} + 3 {H_{2} O}_{(l)} Δ H = - 397.8 k c a l . . . . . (1)$
$2 C_{(s)} + 3 {H_{2}}_{(g)} ⟶ {C_{2} H_{6}}_{(g)} Δ H = - 21.1 k c a l . . . . . (2)$
${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(l)} Δ H = - 94.1 k c a l$
${H_{2} O}_{(l)} ⟶ {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} Δ H = + 94.1 k c a l$
$3 \times [{H_{2} O}_{(l)} ⟶ {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} Δ H = + 94.1 k c a l]$
$3 {H_{2} O}_{(l)} ⟶ 3 {H_{2}}_{(g)} + \frac{3}{2} {O_{2}}_{(g)} Δ H = + 282.3 k c a l . . . . . (3)$
Adding equation $(1), (2) & (3)$ , we get
$2 C_{(s)} + 2 {O_{2}}_{(g)} ⟶ 2 {C O_{2}}_{(g)} Δ H = - 136.6 K c a l$
$\frac{1}{2} \times (2 C_{(s)} + 2 {O_{2}}_{(g)} ⟶ 2 {C O_{2}}_{(g)} Δ H = - 136.6 K c a l]$
$C_{(s)} + {O_{2}}_{(g)} ⟶ {C O_{2}}_{(g)} Δ H = - 68.3 K c a l$
Hence the heat of foemation of $C O_{2}$ will be $- 68.6 k c a l$ .

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When 2 mole of $C_{2} H_{6}$ are completely burnt, 3129 KJ of heat is liberated. Calculate the heat of formation of $C_{2} H_{6} . △ H_{f}$ for $C O_{2} & H_{2} O$ are -395 KJ and -286 KJ respectively.

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Calculate the heat of formation of CO2 by following data.(Δ Hc) C2H6 = −397.8 Kcal(Δ Hf) C2H6 = −21.1 Kcal(Δ Hf) H2O = −94.1 Kcal

Calculate the heat of formation of $C O_{2}$ by following data.
( $Δ$ Hc) $C_{2} H_{6}$ = $- 397.8$ Kcal
( $Δ$ Hf) $C_{2} H_{6}$ = $- 21.1$ Kcal
( $Δ$ Hf) $H_{2} O$ = $- 94.1$ Kcal