Question

8. A typical candy bar has 35.0 g of glucose. The human body generates energy from glucose by the thermochemical equation provided below C6H1206(S) + 6 O2(g) → 6 CO2(g) + 6 H2O(1) AH=-670 kcal How much energy is obtained from the digestion of a typical candy bar? Show your work Final Answer You are provided the following data: H2(g) → 2 H(g) Br(g) → 2 Br(g) H2(g) + Br2(g) → 2 HBr(g) AH = 436.4 kJ/mol AH = 192.5 kJ/mol AH = - 72.4 kJ/mol Calculate AH for the reaction H(g) + Br(g) → HBr(g) Show your work What is AH?

Please write clearly and use sig figs. thanks in advance

Answer 1

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a C6H12O6 (5) + 60₂ (8) - 660₂18) + 6H₂018). DH =-670 kcal. weight of Candy's glucose = 35.0g Molar mass of glucose = 180g 2. Moles of glucose = 35 mol 180 from stoichometry of each mol of glucose reaction we can say gives - 670 kcal energy. i Energy obtained = 180 -670 × 35 mate kcal = |- 130.28 kcal. / Ane g. Ho (g) → 2H (8) DH, = 43604 kJ/mol - o Br, (g) → 2 Br(9) DHE = 192.5 kJ/mol - Hy + Br, → 2 Hbr la) DHz = -7204 kJ/mol -() H18) + Brig) Hbr (g) OH = 9 if we subtract ① & 1 from (11) we get 2H + 2 Br 2 HBr and multiply by 1/2, we will get required ean. SH = † (pHg - SHq -OH) = (-724-19205 = - 350.65 kJ/mol = - 351 kJ/mol