Question

Review Co Part A A calorimeter contains 300 ml. of water at 135 C When 2.10 g of X (a substance with a molar mass of 450 g/mol) is added, it dissolves via the reaction X(s) + H20(1)-+X(aq) and the temperature of the solution increases to 275 °C Calculate the enthalpy change. AH for this reaction per mole of X Assume that the specific heat of the resulting solution is equal to that of water [4.18 /(C) that density of water is 100 g/ml, and that no heat is lost to the calorimeter itself, nor to the surroundings Express the change in enthalpy in kilojoules per mole to three significant figures. View Available in VAL AH- J/mol Subm Part

Answer 1

Given: 30 ml of H₂0 at 13.5% react with 2.10g of Xis) whose molar mass is 450g/mal specific wat of solution = sbecific heat of water ly = 4.18 Density of water, d=1g/mL. final temperature of solution, TA = 27.50 solution: Inital temperature of Soln, Tj= 13.5°c Change in temp, OT = T1-T; = 27.50-13.50 = 14°C Volume a mass of water, mw= density= = 30mix lame mw = 303 Total mass of soln, Moolna mwt mass of solute Since, no heat is lost to the calorimetes itself, nor to 30.09 + 2.10g= 39.13 Therefore, heat released by Xosy = heat gain by saln Heat gain by soln, Q son = msaint eux AT 32.1g * 4.183lg'c * = 1,878.53 Hence, Heat released by 2.10g of X (6) is 1,878.5J Therefore, heat released by 45.g of Xs, will be 1878.55*** Heat released by Imol of ty 1878.55 the sumoundings. x 146 9.10 * 4.5g 9.19 = 40,253.45 = 40.95 KJ Answer: The change in enthol byg= 40.35 kJ/mol

mo b) Griven Reaction: Gatiga On 183+ 130919) - 12009 () * " Hop) mass of CalHggO1 (9) 10.og molar mass of Cia Hagon Mo = 349 g/mol Heat capacity of Calorimeter, C = 7.5 kJ/€ Increase in temperature inside the calorimeter was 22.0° (s)! AT= 22.0°C heat Heat deased by combustion of 10.0g Ca Hago, is equal to gain by calorimeter. formula, CXAT, ce heat capacity Heat gain by calorimeter, Q = EXAT Q = (7.5 kJ/ )*( 22.00) Q = 165 KJ log of Cig Hoa Oine, teleased heat, Q = 16565 mol x 349g since, released heat, Q=1654JX Therefore, 349g (imol) of Ga Harics) log for Imodofciales E Answer: The change in internal energy, AE = 5643kJ/mol Qmos 5643 KJ

Part A ;

The heat released by the reaction will be equal to the heat gain by the solution (water and X(s)).

Therefore, we can calculate the heat released by 2.1 g of X(s). Then find heat for its molar mass that will be the change in the enthalpy of X(s) per mol.

Part (B)

Here, Heat released by the combustion of 10.0 g sucrose is equal to the heat gain by the calorimeter to increase its temperature by 22.0 degrees centigrade.

Therefore we can calculate heat released by 342 g (1 mol) of sucrose = 5643kJ/mol. This is the change in internal energy.