Question

1. a coffee cup calorimeter has c_cal = 105.5 j °c⁻¹ and is initially at 24.69 °c. when a 200.0 ml solution of 0.100 mol l⁻¹ agno₃ at 24.69 °c is mixed with a 100.0 ml solution of 0.100 mol l⁻¹ nacl at 24.69 °c, the temperature of the solution and the coffee cup calorimeter rises to 25.16 °c. assuming that the density of the solution is 1.00 g ml⁻¹ and the heat capacity of the solution is 4.184 j g⁻¹ °c⁻¹, determine the value of δ_rh° (in kj mol⁻¹) for the reaction: nacl(aq) + agno₃(aq) - nano₃(aq) + agcl(s) δ_rh° =? hint: consider all components that are gaining or losing heat in this reaction. a) -59 kj mol⁻¹ b) -5.2 kj mol⁻¹ c) -64 kj mol⁻¹ d) -15 kj mol⁻¹ e) -10 kj mol⁻¹

Explanation:

Step1: Calculate the total mass of the solution

The total volume of the solution is $V = 200.0\ mL+100.0\ mL = 300.0\ mL$. Given the density $
ho = 1.00\ g\ mL^{-1}$, using the formula $m=
ho V$, we have $m = 1.00\ g\ mL^{-1}\times300.0\ mL=300\ g$.

Step2: Calculate the heat absorbed by the solution

Using the formula $q_{sol}=mC_{sol}\Delta T$, where $m = 300\ g$, $C_{sol}=4.184\ J\ g^{-1}\ ^{\circ}C^{-1}$ and $\Delta T=25.16^{\circ}C - 24.69^{\circ}C=0.47^{\circ}C$. So $q_{sol}=300\ g\times4.184\ J\ g^{-1}\ ^{\circ}C^{-1}\times0.47^{\circ}C = 591.924\ J$.

Step3: Calculate the heat absorbed by the calorimeter

Using the formula $q_{cal}=C_{cal}\Delta T$, with $C_{cal}=105.5\ J\ ^{\circ}C^{-1}$ and $\Delta T = 0.47^{\circ}C$. So $q_{cal}=105.5\ J\ ^{\circ}C^{-1}\times0.47^{\circ}C=49.585\ J$.

Step4: Calculate the total heat released by the reaction

The total heat released by the reaction $q_{rxn}=-(q_{sol}+q_{cal})=-(591.924\ J + 49.585\ J)=- 641.509\ J$.

Step5: Calculate the number of moles of reactants

For $AgNO_3$, $n_{AgNO_3}=M\times V=0.100\ mol\ L^{-1}\times0.200\ L = 0.02\ mol$. For $NaCl$, $n_{NaCl}=M\times V=0.100\ mol\ L^{-1}\times0.100\ L=0.01\ mol$. The limiting - reactant is $NaCl$ (since the mole - ratio of $NaCl$ to $AgNO_3$ in the reaction is 1:1).

Step6: Calculate the molar enthalpy change

Using the formula $\Delta_{r}H^{\circ}=\frac{q_{rxn}}{n}$, where $n = 0.01\ mol$ and $q_{rxn}=-641.509\ J$. Converting $q_{rxn}$ to $kJ$ ($q_{rxn}=- 0.641509\ kJ$), we get $\Delta_{r}H^{\circ}=\frac{-0.641509\ kJ}{0.01\ mol}=-64.1509\ kJ\ mol^{-1}\approx - 64\ kJ\ mol^{-1}$.

Answer:

C. $-64\ kJ\ mol^{-1}$