In the lab, I mix two solutions together in a coffee cup in order to perform the

Question

In the lab, I mix two solutions together in a coffee cup in order to perform the following reaction:

Pb(NO3)2(aq) + 2KI(aq) -->PbI2(s) + 2KNO3(aq)

?Hrxn= ?

Solution A is 26.7 mL of 0.342 M Pb(NO3)2 and solution B is 50.0 mL of 0.412 M KI. Both solutions are initially at the same temperature (22.54°C), and after the reaction proceeds the temperature of the solution is 22.67 °C. What is the enthalpy change for this reaction? For this problem, you may assume that the density and specific heat of the solutions are the same as water.

MM Pb(NO3)2= 331.2g/mol

MM KI= 166.0g/mol

MM PbI2 = 461.0 g/mol

MM KNO3 = 101.1 g/mol

Explanation / Answer

Moles of Lead nitrate solution taken = Molarity*Volume = 0.342*0.0267 = 0.0091

Moles of KI taken = 0.412*0.050 = 0.0206

So we see KI is in excess amount than required, so Lead nitrate is the limiting reagent.

Total volume = 26.7+50 = 76.7 mL

Mass of solution = Density*Volume = 76.7*1 = 76.7 g

So, Heat released by reaction is:

Q = m*C*dT = 76.7*4.184*(22.67-22.54) = 41.72 J

So,

dHrxn = Q/moles of limiting reagent = 41.72/0.0091 = 4.584 kJ/mol

Hope this helps !

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