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Heat of Formation

The heat evol...

Question

The heat evolved on combustion of 1 g starch $(C_{6} H_{10} O_{5})_{n}$ into $C O_{2}$ and $H 2 O (l)$ is $17.49 k J g^{- 1}$ . Then the enthalpy of formation of 1 g starch is:

$[G i v e n Δ_{f} H o f H_{2} O (l) = - 285.85 k J m o l^{- 1} . Δ_{f} H o f C O_{2} = - 293.7 k J m o l^{- 1}]$

- 1.22 k J

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- 3.71 k J

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- 4.21 k J

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- 2.21 k J

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Solution

The correct option is D $- 2.21 k J$
Given,
$H_{2} + \frac{1}{2} O_{2} ⟶ H 2 O (l); Δ H = - 285.85 k J m o l^{- 1}$ ...( i)
$C + O_{2} ⟶ C O_{2} (g); Δ H = - 293.7 k J m o l^{- 1}$ ..( ii)

$∴$ 1g starch on combustion gives 17.49 kJ energy.

$∴$ 162 n g starch on combustion gives $= 17.49 \times 162 n k J e n e r g y$

$= 2833.38 n k J$

Molar mass of starch $(C_{6} H_{10} O_{5})_{n} = 162 n$

Therefore,
$(C_{6} H_{10} O_{5})_{n} + 6 n O_{2} ⟶ 6 n C O_{2} + 5 n H_{2} O; Δ H = - 2833.38 n k J$ ...( iii)

Multiply eq. (ii) by 6n and eq. (i) by 5n. then add
$5 n H_{2} + \frac{5 n}{2} O_{2} + 6 n C + 6 n O_{2} ⟶ 5 n H_{2} O + 6 n C O_{2}; Δ H = - 3191.45 n k J$ ...( iv)

Subtract eq. (iii) from eq. (iv):

$6 n C + 5 n H_{2} + \frac{5 n}{2} O_{2} ⟶ (C_{6} H_{10} O_{5})_{n}; Δ H = - 358.07 n k J$

$∴$ Heat of formation for $162 n$ g starch $= - 358.07 n k J$

$∴$ Heat of formation for 1 g starch $= \frac{358.07 n}{162 n} = - 2.21 k J$

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