Question

A 0.500 L sample of 0.200 M NaCl(aq) is added to 0.500 L of 0.200 AgNO3(aq) in a
calorimeter with a known total heat capacity equal to 4.6*103 J/K . The observed $\Delta$ T
is +1.423 K. Calculate the value of $\Delta$ Horxn for the following reaction.

AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)

Answer 1

Moles NaCl = volume x molarity

= 0.5 L x 0.2 M

= 0.1 mol

Similarly Moles AgNO3

=0.5 L x 0.2 M

= 0.1 mol

So, 0.1 mol of AgCl will be form in the reaction.

Density of water = 1 gmL-1

So, total mass of solution = 500 g + 500 = 1000 g

q = q(H2O) + q (calorimeter)

= m x c(water) x dT + c( calorimeter) x dT

Where, (dT) = temperature change

C=specific heat capacity

m = mass of the water

Putting the respective values, we get

q = 1000 g x 4.184 J /g °C x (1.423)°C + (4.6 x 10^3J /°C x (1.423)°C)

q= 5953.83 J + 6545.8 J

q = 12499.63 J

q = 12.499 kJ

Now, delta H(rxn) = 12. 499 kJ / 0.1 mol

= 124.99 kJ /mol