Part A A calorimeter contains 32.0 mL of water at 12.0 oC . When 1 .90 g of X (a

Question

Part A A calorimeter contains 32.0 mL of water at 12.0 oC . When 1 .90 g of X (a substance with a molar mass of 80.0 g/mol ) is added, it dissolves via the reaction x(s) H20(1)->X(aq) and the temperature of the solution increases to 30.0 °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/(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. View Available Hint(s) kJ/mol Submit Part B Consider the reaction C12H22011 (s) 1202 (g-12CO2(g) 11H20(1) in which 10.0 g of sucrose, C12H22011, was burned in a bomb calorimeter with a heat capacity of 7.50 k.J/°C. The temperature increase inside the calorimeter was found to be 22.0° C. Calculate the change in internal energy, 3, for this reaction per mole of sucrose Express the change in internal energy in kilojoules per mole to three significant figures View Available Hint(s) kJ/mol

Explanation / Answer

Part A :

In this problem heat is consumed in increasing the temperature of water

heat Q = M*C* delta T

M = mass of water = 32 ml * 1 gram / 1 ml = 32 gram (density of water is 1.00g/mL )

delta T = final temperature - initial temperature
= 30- 12 = 18 C

specific heat capacity C = 4.18 (J/g(degree celcius)

put the respective value in equation

Q = M*C* delta T

We get,

Heat Q = 32 * 4.18*18
= 2407. 68 J

When 1.90g of X (a substance with a molar mass of 80g/mol ) is added
number of moles of substance X = Mass / molar mass
= 1.90 / 80.0
= 0.0237 mol

Enthalpy change = heat required / number of moles
= 2407.68 j / 0.0237mol
= 101589.8 j/mol

Since 1000 j = 1 kj
convert in kilojoules
Enthalpy change = 101589.8 Jmol-1/ 1000 kj
=101.59 kJmol-1

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