Question

30.0 ml of 2.0 m nh₃(aq) at 21.0°c is mixed with 30.0 ml of 2.0 m hcl(aq) at 21.0°c in a calorimeter. as the reaction proceeds, the temperature rises to 33.56°c. assume the densities of the solutions are both 1.00 g/ml and the specific heats are 4.18 j/g°c. calculate the enthalpy change (δh) of the reaction in kj.
a 3150 kj
b -3.15 kj
c 3.15 kj
d -3150 kj

Explanation:

Step1: Calculate the total mass of the solution

The total volume of the solution is $V = 30.0\ mL+30.0\ mL=60.0\ mL$. Since the density $
ho = 1.00\ g/mL$, using the formula $m=
ho V$, we have $m = 1.00\ g/mL\times60.0\ mL = 60.0\ g$.

Step2: Calculate the heat absorbed by the solution

Use the formula $q = mc\Delta T$, where $c = 4.18\ J/g^{\circ}C$, $\Delta T=T_{final}-T_{initial}=33.56^{\circ}C - 21.0^{\circ}C=12.56^{\circ}C$. Then $q=60.0\ g\times4.18\ J/g^{\circ}C\times12.56^{\circ}C=3150\ J$.

Step3: Determine the number of moles of the reactants

The number of moles of $NH_3$ and $HCl$ is $n = M\times V$, where $M = 2.0\ M$ and $V=0.030\ L$. So $n = 2.0\ mol/L\times0.030\ L = 0.06\ mol$.

Step4: Calculate the enthalpy change of the reaction

The heat released by the reaction is equal to the heat absorbed by the solution in magnitude. Since the reaction is exothermic (temperature rises), $\Delta H=-\frac{q}{n}$. Convert $q = 3150\ J = 3.15\ kJ$. Then $\Delta H=-\frac{3.15\ kJ}{0.06\ mol}=- 3.15\ kJ$ (the negative sign indicates exothermic reaction).

Answer:

B. -3.15 kJ