Calculate enthalpy of formation of D-ribose in Kj/mol. | TestHub

Passage / Comprehension

A sample of the sugar D-ribose (C₅H₁₀O₅) of mass $0.727 g$ was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature rose by $0.910 K$ . In a separate experiment in the same calorimeter, the combustion of $0.825 g$ of benzoic acid, for which the internal energy of combustion is $- 3251 k J m o l^{- 1}$ , gave a temperature rise of $1.940 K$ .

Given:

$Δ_{f} H_{C O_{2}}^{0} = - 393 k J m o l$

$Δ_{f} H_{H_{2} O}^{0} = - 286 k J m o l$

Calculate enthalpy of formation of D-ribose in Kj/mol.

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Answer:

Solution:

Benzoic acid: moles = $0.825/122.1 \approx 0.00676 m o l$ $\Rightarrow$ heat = $0.00676 \times 3251 \approx 21.96 k J$

Calorimeter constant $C = 21.96/1.940 \approx 11.32 k J K^{- 1}$

Ribose heat released = $11.32 \times 0.910 \approx 10.30 k J$

Moles ribose = $0.727/150.13 \approx 0.00484 m o l$ $\Rightarrow$ $Δ U_{co mb} \approx - 10.30/0.00484 \approx - 2128 k J m o l^{- 1}$

C₅H₁₀O₅ + 5O₂ → 5CO₂ + 5H₂O, so $Δ H = Δ U$ ( $Δ n = 0$ )

$5 (- 393) + 5 (- 286) = - 3395 k J m o l^{- 1}$

$- 2128 = - 3395 - Δ H_{f}$ $\Rightarrow$ $Δ H_{f} = - 1267 k J m o l^{- 1}$ $\Rightarrow$ (A) $- 1267.77 k J m o l^{- 1}$

Chemistry - Thermochemistry Question with Solution | TestHub

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