The heat evolved on combustion of 1 g of starch, ( C 6 H 10 O 5 ) x , into CO 2 ( g ) and H 2 O ( 1 ) is 4.6 kcal . Heat of formation of CO 2 ( g ) and H 2 O ( 1 ) are -94.2 and − 68.4

Question

The heat evolved on combustion of 1 g of starch, ( C 6 ​ H 10 ​ O 5 ​ ) x ​ , into CO 2 ​ ( g ) and H 2 ​ O ( 1 ) is 4.6 kcal . Heat of formation of CO 2 ​ ( g ) and H 2 ​ O ( 1 ) are -94.2 and − 68.4 kcal / mol , respectively. The magnitude of standard enthalpy of formation of 1 g of starch (in kcal) is

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To find the standard enthalpy of formation of starch, we use the combustion reaction and Hess's Law.

The combustion of starch can be represented as:

$(C_{6} H_{10} O_{5})_{x} (s) + 6 x O_{2} (g) \to 6 x C O_{2} (g) + 5 x H_{2} O (l)$

The molar mass of the repeating unit $C_{6} H_{10} O_{5}$ is $6 (12) + 10 (1) + 5 (16) = 72 + 10 + 80 = 162 g / m o l$ .

Given:

Heat evolved on combustion of 1 g of starch = 4.6 kcal.

So, $Δ H_{co mb u s t i o n}$ for 1 g of starch = -4.6 kcal.

For 1 mole of the repeating unit (162 g), $Δ H_{co mb u s t i o n} = - 4.6 k c a l / g \times 162 g / m o l = - 745.2 k c a l / m o l$

Using Hess's Law:

$Δ H_{co mb u s t i o n} = \sum Δ H_{f}^{\circ} (p ro d u c t s) - \sum Δ H_{f}^{\circ} (re a c t an t s)$

$Δ H_{co mb u s t i o n} = [6 x Δ H_{f}^{\circ} (C O_{2}) + 5 x Δ H_{f}^{\circ} (H_{2} O)] - [x Δ H_{f}^{\circ} (C_{6} H_{10} O_{5}) + 6 x Δ H_{f}^{\circ} (O_{2})]$

Since $Δ H_{f}^{\circ} (O_{2}) = 0$ , and dividing by $x$ :

$Δ H_{co mb u s t i o n} (p er re p e a t in g u ni t) = [6Δ H_{f}^{\circ} (C O_{2}) + 5Δ H_{f}^{\circ} (H_{2} O)] - Δ H_{f}^{\circ} (C_{6} H_{10} O_{5})$

Substitute the given values:

$Δ H_{f}^{\circ} (C O_{2}) = - 94.2 k c a l / m o l$

$Δ H_{f}^{\circ} (H_{2} O) = - 68.4 k c a l / m o l$

$Δ H_{co mb u s t i o n} (p er re p e a t in g u ni t) = - 745.2 k c a l / m o l$

$- 745.2 = [6 (- 94.2) + 5 (- 68.4)] - Δ H_{f}^{\circ} (C_{6} H_{10} O_{5})$

$- 745.2 = [- 565.2 - 342.0] - Δ H_{f}^{\circ} (C_{6} H_{10} O_{5})$

$- 745.2 = - 907.2 - Δ H_{f}^{\circ} (C_{6} H_{10} O_{5})$

$Δ H_{f}^{\circ} (C_{6} H_{10} O_{5}) = - 907.2 + 745.2$

$Δ H_{f}^{\circ} (C_{6} H_{10} O_{5}) = - 162.0 k c a l / m o l$

This is the enthalpy of formation for 1 mole (162 g) of the repeating unit.

To find the enthalpy of formation for 1 g of starch:

$Δ H_{f}^{\circ} (s t a rc h p er g) = \frac{- 162.0 k c a l / m o l}{162 g / m o l} = - 1.0 k c a l / g$

The magnitude of the standard enthalpy of formation of 1 g of starch is 1.0 kcal.

Final Answer: 1

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