The standard enthalpies of formation of C O 2(g), H 2 O (l) and glucose (s) at 25° C are -400 kJ / mol ,-300 kJ / mol and -1300 kJ / mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° C is

Question

The standard enthalpies of formation of C O 2(g), H 2 O (l) and glucose (s) at 25° C are -400 kJ / mol ,-300 kJ / mol and -1300 kJ / mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° C is (A) + 2900 kJ (B) - 2900 kJ (C) - 16.11 kJ (D) + 16.11 kJ

Tardigrade Admin · Accepted Answer

PLAN ΔC H° (Standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised. Also standard heat of formation (Δf H°) can be taken as the standard of that substance. H CO 2°=Δf H°( CO 2)=-400 kJ mol -1 H H 2 O °=Δf H°( H 2 O )=-300 kJ mol -1 H text glucose °=Δf H°( glucose )=-1300 kJ mol -1 H O 2°=Δf H°( O 2)=0.00 C 6 H 12 O 6(s)+6 O 2(g) longrightarrow 6 CO 2(g)+6 H 2 O (l) Δc H°( glucose )=6[Δf H°( CO 2)+Δf H°( H 2 O )]-[Δf H°( C 6 H 12 O 6)+6 Δf H°( O 2)] =6[-400-300]-[-1300+6 × 0] =-2900 kJ mol -1 Molar mass of C 6 H 12 O 6=180 g mol -1 Thus, standard heat of combustion of glucose per gram =(-2900/180)=-16.11 kJ g -1 To solve such problem, students are advised to keep much importance in unit conversion. As here, value of R (8.314 J K -1 mol -1) in JK -1 mol -1 must be converted into kJ by dividing the unit by 1000 .

Q. The standard enthalpies of formation of CO2​(g),H2​O(l) and glucose (s) at 25∘C are −400kJ/mol,−300kJ/mol and −1300kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25∘C is

+ 2900 kJ

- 2900 kJ

- 16.11 kJ

+ 16.11 kJ

Q. The standard enthalpies of formation of $C O_{2} (g), H_{2} O (l)$ and glucose $(s)$ at $2 5^{\circ} C$ are $- 400 k J / m o l, - 300 k J / m o l$ and $- 1300 k J / m o l$ , respectively. The standard enthalpy of combustion per gram of glucose at $2 5^{\circ} C$ is