Calculation for Calorimetry
CALCULATIONS Determining the amount Limiting Reagent used. nlimiting reagent = Molarity x Volume or Mass / Molar Mass
Example:
Limiting reagent is 5mL of 1.0 M HCl nlimiting reagent = Molarity x Volume nlimiting reagent = (1.0 [mol/L]) x 0.005 [L])
= 0.005 mol
Determining the qrxn and qcal. qrxn + qcal = 0
-qrxn = qcal qrxn = ΔHrxn x nlimiting reagent qcal = Ccal ΔT qrxn = - Ccal ΔT + mcsolid ΔT (note: only if there is a precipitate formed in the reaction)
Examples:
(1) Calibration of the calorimeter given that:
ΔHrxn = -55.8 kJ/mol and nLR = 0.005 mol qrxn = ΔHrxn x nlimiting reagent qrxn = -55.8 [kJ/mol] x 0.005 [mol] = -279 J qcal = -(219 J) = 279 J
(2) Determining the qrxn of a given chemical reaction: NH3 (aq) + H+ (aq) NH4+ (aq)
And given that: ΔT = 3.5 °C and Ccal=111.6 J/°C qrxn = - Ccal ΔT + mcsolid ΔT qrxn = -( 111.6 [kJ/°C] x 3.5 [°C]) = -390.6 J qcal = -(-390.6 J) = 390.6
Determining the Ccal.
Ccal = qcal / ΔT
Example:
Given qrxn = -279 J and ΔT = 2.5 °C
Ccal = -qrxn / ΔT
Ccal = -(-279 J) / (2.5 °C) = 111.6 J/°C
Determining the experimental ΔHrxn.
ΔHrxn = qrxn / nLR
Example:
Given: NH3 (aq) + H+ (aq) NH4+ (aq)
With qrxn = -390.6 J and nLR = 0.005 mol
ΔHrxn = qrxn / nLR
ΔHrxn = -390.6 J / 0.005 mol = -78.1 kJ/mol
Determining the theoretical ΔHrxn.
ΔHrxn = ∑nproductH°f product - ∑nreactantH°f reactant
Example:
Given that: NH3 (aq) + H+ (aq) NH4+ (aq) Substance ΔH°f (kJ/mol)
NH3 (aq) -80.29
H+ (aq) 0.00
NH4+ (aq) -132.51
ΔHrxn = ∑nproductH°f product - ∑nreactantH°f reactant
ΔHrxn = {-132.51 kJ/mol}-{-80.29 kJ/mol} =
ΔHrxn = -52.2 kJ/mol
Determining the %error.
%error = (|ΔHexperimental – ΔHtheoretical|) / (ΔHtheoretical) x 100%
Example:
Given: ΔHexperimental = -78.1 kJ/mol and ΔHtheoretical = -52.2 kJ/mol
%error = |(ΔHexperimental – ΔHtheoretical) / (ΔHtheoretical) | x 100%
%error = |(-78.1 kJ/mol) – (-52.2 kJ/mol) / -52.2 kJ/mol| x 100% =
Example:
Limiting reagent is 5mL of 1.0 M HCl nlimiting reagent = Molarity x Volume nlimiting reagent = (1.0 [mol/L]) x 0.005 [L])
= 0.005 mol
Determining the qrxn and qcal. qrxn + qcal = 0
-qrxn = qcal qrxn = ΔHrxn x nlimiting reagent qcal = Ccal ΔT qrxn = - Ccal ΔT + mcsolid ΔT (note: only if there is a precipitate formed in the reaction)
Examples:
(1) Calibration of the calorimeter given that:
ΔHrxn = -55.8 kJ/mol and nLR = 0.005 mol qrxn = ΔHrxn x nlimiting reagent qrxn = -55.8 [kJ/mol] x 0.005 [mol] = -279 J qcal = -(219 J) = 279 J
(2) Determining the qrxn of a given chemical reaction: NH3 (aq) + H+ (aq) NH4+ (aq)
And given that: ΔT = 3.5 °C and Ccal=111.6 J/°C qrxn = - Ccal ΔT + mcsolid ΔT qrxn = -( 111.6 [kJ/°C] x 3.5 [°C]) = -390.6 J qcal = -(-390.6 J) = 390.6
Determining the Ccal.
Ccal = qcal / ΔT
Example:
Given qrxn = -279 J and ΔT = 2.5 °C
Ccal = -qrxn / ΔT
Ccal = -(-279 J) / (2.5 °C) = 111.6 J/°C
Determining the experimental ΔHrxn.
ΔHrxn = qrxn / nLR
Example:
Given: NH3 (aq) + H+ (aq) NH4+ (aq)
With qrxn = -390.6 J and nLR = 0.005 mol
ΔHrxn = qrxn / nLR
ΔHrxn = -390.6 J / 0.005 mol = -78.1 kJ/mol
Determining the theoretical ΔHrxn.
ΔHrxn = ∑nproductH°f product - ∑nreactantH°f reactant
Example:
Given that: NH3 (aq) + H+ (aq) NH4+ (aq) Substance ΔH°f (kJ/mol)
NH3 (aq) -80.29
H+ (aq) 0.00
NH4+ (aq) -132.51
ΔHrxn = ∑nproductH°f product - ∑nreactantH°f reactant
ΔHrxn = {-132.51 kJ/mol}-{-80.29 kJ/mol} =
ΔHrxn = -52.2 kJ/mol
Determining the %error.
%error = (|ΔHexperimental – ΔHtheoretical|) / (ΔHtheoretical) x 100%
Example:
Given: ΔHexperimental = -78.1 kJ/mol and ΔHtheoretical = -52.2 kJ/mol
%error = |(ΔHexperimental – ΔHtheoretical) / (ΔHtheoretical) | x 100%
%error = |(-78.1 kJ/mol) – (-52.2 kJ/mol) / -52.2 kJ/mol| x 100% =