Part A A calorimeter contains 20.0 mL of water at 12.5 C . When 1.50 g of X (a s

Question

Part A

A calorimeter contains 20.0 mL of water at 12.5 C . When 1.50 g of X (a substance with a molar mass of 58.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)X(aq) and the temperature of the solution increases to 26.5 C . Calculate the enthalpy change, H, 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/(gC)], 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, E, for this reaction per mole of sucrose. Express the change in internal energy in kilojoules per mole to three significant figures. E = kJ/mol SubmitHintsMy AnswersGive UpReview Part

Explanation / Answer

Part A

mass of water = d*v = 1*20 = 20 grams

specific heat of water = 4.18 J/(gC

DT = 26.5-12.5 = 14 c

qreleased = m*s*DT

= 20*4.18*14

   = 1170.4 joule

   = 1.1704 kj

No of moles of X = w/mwt   = 1.5/58 = 0.026 mole

so that

0.026 mole of X    = 1.1704 kj

1 mole X    = 1.1704/0.026 = 45.0154 kj/mol

answer :

DH rxn = - 45.0 kj/mol

part B

C12H22O11(s)+12O2(g) ---> 12CO2(g)+11H2O(l)

No of moles of sucrose = 10/342.2965 = 0.029 mole

Molar mass of sucrose = 342.2965 g/mol

DT = 22 c

heat capacity = 7.50 kJ/C

heat reased (q) = Cp*DT = 7.5*22 = 165 kj

so that,

0.029 mole of sucrose = 165 kj

1 mole of sucrose = 165/0.029 = 5689.65 kj/mol

DE = - 5689.65 kj/mol

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