Answer:
ΔH = -98 kJ/mol
Explanation:
To calculate the heat change of the reaction:
HCl(aq) + NH₃(aq) → NH₄Cl(aq)
0.2M 0.2M ΔT=2.34°C
1x10²mL 1x10²mL
We need to use the next equation:
[tex] q = mc \Delta T [/tex] (1)
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 {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!
