Question

A 1.67-g sample of solid silver reacted in excess chlorine gas to give a2.21-g sample of pure solid Agcl.The heat given off in this reaction was 1.96 kJ at constant pressure.Given this information,what is the enthalpy of formation of AgCl(s)

Answer 1

Given:

Mass of Ag = 1.67 g

Mass of Cl = 2.21 g

Heat evolved = 1.96 kJ

To determine:

The enthalpy of formation of AgCl(s)

Explanation:

The reaction is:

2Ag(s) + Cl2(g) → 2AgCl(s)

Calculate the moles of Ag and Cl from the given masses

Atomic mass of Ag = 108 g/mol

# moles of Ag = 1.67/108 = 0.0155 moles

Atomic mass of Cl = 35 g/mol

# moles of Cl = 2.21/35 = 0.0631 moles

Since moles of Ag << moles of Cl, silver is the limiting reagent.

Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles

Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol

Ans: Formation enthalpy = 126.5 kJ/mol

Answer 2

Answer: The enthalpy of formation of AgCl is -127.3 kJ/mol

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of silver chloride = 2.21 g

Molar mass of silver chloride = 143.32 g/mol

Putting values in above equation, we get:

$\text{Moles of silver chloride}=\frac{2.21g}{143.32g/mol}=0.0154mol$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the enthalpy change of formation of AgCl, we use the equation:

$\Delta H_f_{AgCl}=\frac{q}{n}$

where,

$q$ = amount of heat released = -1.96 kJ

n = number of moles of AgCl = 0.0154 moles

$\Delta H_f_{AgCl}$ = enthalpy of formation of silver chloride

Putting values in above equation, we get:

$\Delta H_f_{AgCl}=\frac{-1.96kJ}{0.0154mol}=127.3kJ/mol$

Hence, the enthalpy of formation of AgCl is -127.3 kJ/mol