Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (C_cal = 15.5 J °C^-1). When a 100.0 mL volume of 0.100 mol L^-1 AgNO₃ solution is mixed with a 100.0 mL sample of 0.200 mol L^-1 NaCl solution, the temperature in the calorimeter rises to 25.30 °C. Determine the Δ_rH° for the reaction as written below. Assume that the density and specific heat capacity of the solutions is the same as that of water. NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq) Δ_rH° = ?

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat absorbed by the solution. We can use the formula:

q = mCΔT

Where q is the amount of heat absorbed, m is the mass of the solution (which we assume to be 100 g for both solutions), C is the specific heat capacity of the solution (which we assume to be the same as water, 4.184 J/g°C), and ΔT is the change in temperature (which is 0.7°C).

q = (100 g)(4.184 J/g°C)(0.7°C) = 292.88 J

Next, we need to calculate the number of moles of each reactant.

For NaCl:

n = C x V = (0.2 mol/L)(0.1 L) = 0.02 mol

For AgNO3:

n = C x V = (0.1 mol/L)(0.1 L) = 0.01 mol

Since the stoichiometry of the balanced equation is 1:1 for both NaCl and AgNO3, we can see that AgNO3 is the limiting reactant. Therefore, the reaction releases 0.01 mol of heat:

ΔH = -q/n = -(292.88 J)/(0.01 mol) = -29,288 J/mol = -29.3 kJ/mol

Finally, we can use the equation:

ΔG = ΔH - TΔS

Where ΔG is the change in Gibbs free energy, and TΔS is the energy that is unavailable to do work (in this case, due to the increase in entropy of the system). We want to solve for ΔS, so we can rearrange the equation:

ΔS = (ΔH - ΔG)/T

From the balanced equation, we know that ΔS is positive, since there are more moles of products than reactants. Therefore, we can assume that ΔS is equal to the positive value of the entropy change for the balanced equation.

Since ΔG = 0 at equilibrium, we can set ΔG = 0 and solve for ΔS:

ΔS = (ΔH - ΔG)/T = ΔH/T

Plugging in the values:

ΔS = (-29.3 kJ/mol)/(298 K) = -98.3 J/K/mol

Converting to kilojoules:

ΔS = -0.0983 kJ/K/mol

Therefore, the answer is B, -59 kJ/mol.

Two Solutions, Initially at 24