PART A A calorimeter contains 22.0 mL of water at 12.5 C . When 1.30 g of X (a s
ID: 891760 • Letter: P
Question
PART A
A calorimeter contains 22.0 mL of water at 12.5 C . When 1.30 g of X (a substance with a molar mass of 49.0 g/mol ) is added, it dissolves via the reaction
X(s)+H2O(l)X(aq)
and the temperature of the solution increases to 29.0 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/(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.
Explanation / Answer
A).
Given:
Volume of water =22.0 mL
Initial temperature = 12.5 deg C
Mass of X = 1.30
Molar mass of X = 49.0 g per mol
Reaction:
X (s) + H2O (l) --- > X (aq)
Delta T = 29.0 deg
Solution:
We know
q = C m Delta T
C is specific heat , m is mass , Delta T
q is heat absorbed / evolved.
Calculation of Heat gained by water
q = 22.0 mL x (1.00 g /mL) x 4.184 J (g deg C)-1x 29.0 deg C
=2669.392 J
Since it is the heat given by substance so the value of the reaction is having value –ve
q (reaction) = - 2669.392 J
Delta H= q / n
n = number of moles
n of substance = mass / molar mas
= 1.30 g / 49.0 g per mol
=0.026531 mol
Delta H = -2669.392 J / 0.026531 mol
=-1000615.5 J/mol
Delta H in kJ
=- 1000.6 kJ/mol
Part B
C12H22O11(s)+12O2(g)12CO2(g)+11H2O(l)
Given :
Mass of sucrose = 10.0 g ,
Capacity of calorimeter = 7.50 kJ/deg C
Rise in T = 22.0 deg C
We use
q reaction = - q cal
q cal = q(bomb) + q(water )
q (bomb) = Heat capacity x change in T
=7.50 kJ/deg C x 22.0 deg C
=165 kJ
= 165000 J
q water = 10.0 g x 4.184 J (deg C g )-1 x 22.0 deg C
=920.48 J
q cal = 165000 J + 920.48 J
= 165920.5 J
q reaction = -q
= - 165920.5 J
Delta E = q
Since the process is at constant volume
So the Delta E = q
Delta E = - 165920.5 J
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