If enthalpy of combustion of carbon, hydrogen and $C_{3} H_{8}$ are $x_{1}, x_{2}$ and $x_{3}$ per ${m o l}^{- 1}$ respectively, then the enthalpy of formation of $C_{3} H_{8}$ will be

x_{3} - x_{1} - x_{2}

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x_{3} - 3 x_{1} + 4 x_{2}

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3 x_{1} + 4 x_{2} - x_{3}

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x_{1} + x_{2} - x_{3}

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Solution

The correct option is C $3 x_{1} + 4 x_{2} - x_{3}$
$C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 {C O}_{2} (g) + 4 H_{2} O (l); Δ H = x_{3} . . . . (i)$
$C (s) + O_{2} (g) ⟶ {C O}_{2} (g); Δ H = x_{1} . . . . . . . (i i)$
$H_{2} (g) + \frac{!}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = x_{2} . . . . . . (i i i)$
Required reaction is
$3 C (s) + 4 H_{2} (g) ⟶ C_{3} H_{8} (g); Δ H = ?$
By $3 \times E q . (i i) + 4 \times E q . (i i i) - E q . (i)$ . we get
$3 C + 4 H_{2} - C_{3} H_{8} ⟶ 0; Δ H = 3 x_{1} + 4 x_{2} - x_{3}$
$3 C + 4 H_{2} ⟶ C_{3} H_{8}; Δ H = 3 x_{1} + 4 x_{2} - x_{3}$

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If enthalpy of combustion of carbon, hydrogen and C3H8 are x1,x2 and x3 per mol−1 respectively, then the enthalpy of formation of C3H8 will be

If enthalpy of combustion of carbon, hydrogen and $C_{3} H_{8}$ are $x_{1}, x_{2}$ and $x_{3}$ per ${m o l}^{- 1}$ respectively, then the enthalpy of formation of $C_{3} H_{8}$ will be