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Question
Buffy carries out a neutralization reaction in her styrofoam cup calorimeter, with a heat capacity of 41 J/C. She mixes 25.0mL of 3.0M acid solution and 25.0mL of 3.0M base solution at 22.3 degree C. The temperature rose to 36.5 C. D=1.03 g/ml and Heat Capacity= 4.05 J/g*C
-Calculate heat energy absorbed by solution (qsolution)
-Calculate heat energy absorbed by the calorimeter (qcalorimeter)
-Calculate q reaction
-calculate number of moles of acid reacting
-calculate delta H neut. of the reaction in joules/mole
Buffy carries out a neutralization reaction in her styrofoam cup calorimeter, with a heat capacity of 41 J/C. She mixes 25.0mL of 3.0M acid solution and 25.0mL of 3.0M base solution at 22.3 degree C. The temperature rose to 36.5 C.
Heat capacity of Styrofoam cup calorimeter = 41 J/C
Initial T = 22.3 deg C , Volume of acid solution = 25.0 mL and molarity = 3.0 M , molarity of base = 3.0 M and volume is 25.0 mL
Final T = 36.5 deg C
Calculate heat energy absorbed by solution (qsolution)
Step 1) Calculation of mass of solution
Mass of solution = Volume of solution * density
= (25+25 )mL * 1.0 g/mL (density of water = 1.0 g / mL )
= 50.0 g
q = c m delta T
= 4.18 J / g Deg C * 50.0 g * ( 36.5- 22.3 ) deg C
= 2967.8 J
-Calculate heat energy absorbed by the calorimeter (qcalorimeter)
q = sp heat of Calorimeter * T change
= 41 J/ deg C * ( 36.5- 22.3 ) deg C
= 582.0 J
-Calculate q reaction
q = Heat absorbed by solution+ heat absorbed by calorimeter
= 2967.8 J + 582.0 J = 3.55 E 3 J
-calculate number of moles of acid reacting
We calculate number of moles of acid present in the solution by using molarity and volume in L
Number of moles = volume in L * molarity
= 0.025 L * 0.30 M
= 0.0075 mol
Number of moles of base = 0.025 L * 0.3 M = 0.0075 mol
So equal moles of base and acid so 0.0075 mol of acid is reacted
calculate delta H neut. of the reaction in joules/mol
Number of moles of acid = 0.0075 mol
Delta H = - q * n =- 3.55 E 3 J / 0.0075 mol
= - 4.73 E 5 J per mol