A 1.00 x 102mL sample of 0.200 M aqueous hydrochloric acid is added to 1.00 x 102mL of 0.200 M aqueous ammonia in a constant-pressure calorimeter of negligible heat capacity. The following reaction occurs when the two solutionsare mixedHCl(aq)+ NH3(aq)--->NH4Cl(aq)The temperature increase is 2.34°C. Calculate heat change of the reaction per mole of HCl reacted.Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively)

where q: the amount of heat energy lost or gained, m: the mass of the substance, c: the specific heat capacity of the substance and ΔT: the change in temperature of the substance

Assuming that the densities of the solutions are the same as for water, we can determine the mass of the solution:

[tex] d = \frac {m}{V} [/tex]

where d: density, m: mass and V: volume of solution = 100 + 100 = 200mL

[tex] m = d \cdot V = 1 \frac {g}{mL} \cdot 200mL = 200g [/tex]

Now, using the calculated mass in equation (1), and assuming that the specific heats of the solutions are the same as for water, we can find heat change of the reaction:

[tex] q = 200g \cdot 4.184 \frac {J}{g \cdot ^{\circ}C} \cdot 2.34^{\circ}C[/tex]

[tex] q = 1.96 \cdot 10^{3}J [/tex]

This heat is negative because is the heat lost by the reacting HCl and NH₃ and gained by the water, so:

[tex] q = - 1.96 \cdot 10^{3}J [/tex]

To calculate the heat change of the reaction per mole of HCl, we need to divide the heat change by the number of moles, which is called the enthalpy of reaction:

[tex] \Delta H = \frac {q}{moles} [/tex]

[tex] \Delta H = \frac {-1.96 \cdot 10^{3} J}{0.2 \frac{mol}{L} \cdot 0.1L} [/tex]

[tex] \Delta H = -98 \cdot 10^{3} \frac{J}{mol} = -98 \frac {kJ}{mol} [/tex]

So, the heat change of the reaction per mole of HCl reacted, often called enthalpy of reaction, is ΔH = -98 kJ/mol.

Have a nice day!