Calculate change in heat per mole of HCl and NH3 reacted?

Question

A 100. mL sample of 0.200M aq hydrochloric acid is added to 100. mL of 0.200 mL aq ammonia in a calorimeter whose heat capacity(excluding any water) is 480 J/K. The following reaction occurs when the two solutions are mixed.

HCl(aq) + NH3(aq) --> NH4Cl(aq)

The temperature increase is 2.34C. Calculate change in heat per mole of HCl and NH3 reacted?

Answer 1

Step 1: Find the number of moles that you have of HCl and NH3.
.1 L * .2 mol/L = .02 moles HCl and NH3, which will react to form .02 moles of NH4Cl.

Step 2: Find the total amount of heat released.
Heat capacty = 480 J/K
Change in temp = 2.34 C (or 2.34 K)
Heat released = H. capacity * ΔT = 480 * 2.34 = 1123.2 J

Step 3: Given the Heat released and # of moles, find ΔH/mol.
Change in Heat/mole = ΔH / # of moles
1123.2 / .02 = 56160 J/mol

Step 4: Convert to significant digits (3) = 56.2 kJ/mol (solution!)

Answer 2

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