Question

Part A: A calorimeter contains 18.0 mL of water at 13.5∘C. When 2.30 g of X (a substance with a molar mass of 69.0 g/mol) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 30.0∘C. Calculate the enthalpy change, ΔH, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings. Part B: the reaction C12H22O11(s)+12O2(g)→12CO2(g)+11H2O(l) in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/∘C. The temperature increase inside the calorimeter was found to be 22.0∘C. Calculate the change in internal energy, ΔE, for this reaction per mole of sucrose.

Answer 1

Part A

18 ml water = 18 gm (density of water = 1gm/ml) 18 gm + 2.30 gm = 20.30 gm

T = 30 - 13.5 = 16.5 ⁰ C

q = mass of solution specific heat of H₂O_(l)T

= 20.30 4.18 16.5

q = 1400.091 J

molar mass of X = 69gm/mole then 2.30 gm = 2.30/69 = 0.03333 mole

ΔH for 0.03333 mole is 1400.091 then ΔH for 1 mole = 1400.091/0.03333 = 42006.93 J

enthalpy change, ΔH, for this reaction per mole of X = 42006.93 J/mol = 42.00693 KJ/mol

Part B

Heat produced in bomb calorimeter when combusting a given amount of known substance is the internal energy of the substance combusted per the number of mole of the substance that was burned.

Here 10.0 gm of sucrose/ 342.3 gm/mol = 0.02921 mol

you have given heat capacity of calorimeter is 7.50 KJ/⁰C this mean that the tempreture of calorimeter increases

1⁰C when 7.50 KJ of heat is absorbed by the calorimeter from combution process.

since tempreture increases was 22⁰C when combusting the 0.02921 mole of sucrose then the combution process must given off (22.0⁰C) (7.50KJ/⁰C = 165KJ

threfore the change in internal energy per mole of sucrose combusted would be = 165KJ/0.02921 mole =

5649 KJ/mol