# Calculation for Calorimetry

Topics: Mole, Product, Chemical reactions Pages: 2 (340 words) Published: November 24, 2012
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)

SubstanceΔH°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% =...