A calorimeter contains 20.0 mL
A calorimeter contains 20.0 mL of water at 12.5 ∘C . When 1.40 gof X (a substance with a molar mass of 64.0 g/mol ) is added, itdissolves via the reaction
X(s)+H2O(l)→X(aq)
and the temperature of the solution increases to 28.5 ∘C .
Calculate the enthalpy change, ΔH, for this reactionper mole of X.
Assume that the specific heat of the resulting solution is equalto that of water [4.18 J/(g⋅∘C)], that density of water is 1.00g/mL, and that no heat is lost to the calorimeter itself, nor tothe surroundings.
Express the change in enthalpy in kilojoules per mole tothree significant figures.
Consider the reaction
C12H22O11(s)+12O2(g)→12CO2(g)+11H2O(l)
in which 10.0 g of sucrose, C12H22O11, was burned in a bombcalorimeter with a heat capacity of 7.50 kJ/∘C. The temperatureincrease inside the calorimeter was found to be 22.0 ∘C. Calculatethe change in internal energy, ΔE, for this reaction permole of sucrose.
Express the change in internal energy in kilojoules per mole tothree significant figures.
Answer:
1)
weare given the following valuesMass of reactant= 1.40gmoles of reactant = m / molar mass = 1.40 / 64 = 0.0218 molesMass of solution (m ) = 20gIntital temperature T1 = 12.5 °CFinal temperature T2 = 28.5°CChange in temperature ∆T = T2- T1=28.5 – 12.5 ==16 °CSpecific heat capacity of water Cp =4.184 J/g°CHeat generatedQ = m x Cp x ∆T ……….. (1)Plug the values in this formula we getQ = 20 x 4.184 x 16 jQ= 1338.88 jDivide by 1000 to convert in Kj we getQ= 1.339 kjChange in enthalpy∆H = Q / number of moles∆H = 1.339 kj/ 0.0218 mol∆H = 61.4 Kj/ molThis reaction is producing energy so it a exothermic reactionAnd in exothermic reaction ∆H is always negativeSo our answer will ∆H = -61.4 Kj/ mol
2)
The heat produced in a bomb calorimeter when combusting a givenamount of known substance is the internal energy of the substancecombusted per the number of moles of the substance that wasburned.Here 10.0 g of sucrose is (10.0 g of sucrose) / (342.3 g/mol) =0.02921 moles of sucroseYou are given that the heat capacity of the calorimeter is 7.50 kJ/deg C. This means that the temperature of calorimeter increases 1deg C when 7.50 kJ of heat is absorbed by the calorimeter from thecombustion process. Since the temperature increase was 22.0 deg Cwhen combusting the 0.02921 moles of sucrose, then the combustionprocess must have given off (22.0 deg C)(7.50 kJ/deg C) = 165.0kJ..Therefore, the change in internal energy per mole of sucrosecombusted would be = (165.0 kJ) / (0.02921 moles of sucrosecombusted) = 5649 kJ/mol of sucrose combusted.
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