Question

Part A A calorimeter contains 21.0 mL of water at 11.0 ∘C . When 1.60 g...

Part A A calorimeter contains 21.0 mL of water at 11.0 ∘C . When 1.60 g of X (a substance with a molar mass of 72.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 26.0 ∘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. Express the change in enthalpy in kilojoules per mole to three significant figures. part b Consider 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. Express the change in internal energy in kilojoules per mole to three significant figures.

Homework Answers

Answer #1

PART A

mass of the mixture = v*d+mass of X

= 21.0*1+1.6 = 22.6 g

s = specific heat of mixture = 4.184 j/g.c

DT = 26-11 = 15c

heat released(q) = m*s*DT

     = 22.6*4.184*15

     = 1.418 kj

no of mol of X = w/mwt = 1.6/72 = 0.022 mol

DHrxn = -q/n

       = -1.418/0.022

       = -64.4 kj/mol

part B

heat released during combustion reaction(q) = C*DT

    = 7.5*22

     = 165 kj

No of mol of sucrose burned = w/mwt = 10/342.3 = 0.0292 mol

DUrxn = -q/n = -165/0.0292 = -5650.7 kj/mol

           answer: -5.65*10^3 kj/mol

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