..........................................................................HCl –
ID: 990843 • Letter: #
Question
..........................................................................HCl – NaOH...........HNO3-NaOH
Initial Temperature from graph, oC...........................23.2......................24.1
Final Temperature from graph, oC............................36.75....................36.45
Temperature change, oC..........................................13.45....................12.35
Heat gained by solution, J.........................................13.6.....................12.4
Mass of solution, g....................................................104.4531.............94.5181
Specific Heat, J / g oC...............................................4.184...................4.184
...................................................................................NaCl...................NaNO3
Heat gained by the water solution in the surroundings..?.....................?
Heat capacity of calorimeter, J/oC...............................0.615 J/C..........0.615J/C
Heat gained by calorimeter, J.......................................?.......................?
Heat of neutralization, J...............................................?........................?
Moles of water formed in reaction................................?........................?
Molar enthalpy of neutralization...................................?........................?
Show a sample of each calculation performed. This calculation should include the math formula and the example with substituted values. All data should have proper significant figures and units. we placed 50 ml of 2.00M HCL into the inner cup. stired the solution untill the temperatew stabilized. placed 50 ml of 2.02 M NaOH into 100 m beake. pour the NaOH solution as completely as possible into the calorimeter with the first solution. we also did the same proces using 50 ml 2.00 M HNO3
Explanation / Answer
Solution :-
Calculation for the HCl-NaOH
Heat gained by solution in the surrounding
q= m*s*delta T
q= 104.4531 g * 4.184 J per g C * 13.45 C
= 5883 J
Heat gained by calorimeter
q cal = s cal * delta T
= 0.615 J per degree C * 13.45 C
= 8.27 J
Heat of neutralization = 5883 J + 8.27 J = 5891 J
Since the temperature is rising means reaction is exothermic so the sign is negative so heat of neutralization = -5891 J
HCl and NaOH have 1 :1 mole ratio and the volumes of solution used are same but the molarity of the NaOH is higher so the limiting reactant is HCl
So the moles of water formed are same as moles of HCl reacted
Moles of HCl = molarity * volume in liter
= 2.00 mol per L * 0.050 L
= 0.100 mol HCl
So the moles of water formed = 0.100 mol
Molar enthalpy of neutralization = heat of neutralization / moles of water
= -5891 J / 0.100 mol
= -58910 J/mol
Lets convert it to kJ
-58910 J per mol * 1 kJ / 1000 J = -58.91 kJ/mol
So the molar heat of neutralization is -58.91 kJ/mol
Now calculation for the HNO3- NaOH
Heat gained by water solution in the surrounding
q = m*s*delta T
= 94.5181 g * 4.184 J per g C * 12.4 C
= 4904 J
Heat gained by calorimeter = q = S cal * delta T
= 0.615 J per degree C * 12.4 C
= 7.63 J
Heat of neutralization = 4904 J + 7.63 J = 4911.63 J
Reaction is exothermic so the sign is negative so its - 4911.63 J or we can round it to -4912 J
Moles of water formed are same as moles of HNO3 reacted
So the moles of HNO3 = molarity * volume in liter
= 2.00 mol per L * 0.05 L
=0.100 mol HNO3
So the moles of water formed = 0.100 mol
Molar heat of neutralization = heat of neutralization / moles of water
= -4912 J / 0.100 mol
= -49120 J /mol
Lets convert it to kJ
-49120 J per mol * 1 kJ / 1000 J = -49.12 kJ/mol
So the molar heat of neutralization is -49.12 kJ/mol
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