Question

When a solutions of NaI and AgNO₃ are combined, solid silver iodide is produced accroding to the reaction below.

NaI (aq) + AgNO₃ (aq) → AgI (s) + NaNO₃ (aq)             ΔH = ?

102.8 ml of 0.634 M NaI are combined with 102.8 ml of 0.634 M AgNO₃ in a perfect calorimeter. Both of the starting solutions were initially at a temperature of 33.63 °C. After the reaction is complete the final temperature of the solution in the calorimeter is 41.96. (You may assume that the specific heats of all solutions are the same as that of pure water and that the denisty of each solution is 1.00 g/ml.)

Calculate the value of the enthalpy change (ΔH) for the reaction shown above (in kJ).

Answer kJ

Answer 1

mol of reactant reacting = M*V

= 0.634 M * 0.1028 L

= 0.0652 mol

0.0652 mol of each reactant reacts

volume of solution = 102.8 mL + 102.8 mL

= 205.6 mL

Since density of solution is 1 g/mL,

mass of solution = 205.6 g

Heat absorbed by solution,

Q = m*C*delta T

= 205.6 g * 4.184 J/g.oC * (41.96 - 33.63) oC

= 7166 J

This is heat released by reaction

So, for reaction,

Q = -7166 J = -7.166 KJ

n = 0.06522 mol

Now use:

delta H = Q/n

= -7.166 KJ / 0.06522 mol

= -110. KJ/mol

Answer: -110. KJ/mol