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Chapter 6: Problem 68

In a coffee-cup calorimeter, 1.60 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with 75.0 \(\mathrm{g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\) . After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\) . Assuming the solution has a heat capacity of 4.18 $\mathrm{J} / \mathrm{C} \cdot \mathrm{g}$ and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of kJ/mol.

Short Answer

Expert verified
The enthalpy change for the dissolution of \(\mathrm{NH}_{4}\mathrm{NO}_{3}\) in the given coffee-cup calorimeter experiment is \(-26.02 \ \mathrm{kJ/mol}\).

Step by step solution

01

Calculate the heat change (q) of the solution

To calculate the heat change, we will use the formula: q = mcΔT where q represents the heat change or transferred energy, m is the mass of the solution, c is the heat capacity of the solution, and ΔT is the change in temperature. Given: Mass of water (m): 75.0 g Heat capacity of the solution (c): 4.18 J/g·°C Initial temperature (Ti): 25.00 °C Final temperature (Tf): 23.34 °C First, let's find the change in temperature (ΔT): ΔT = Tf - Ti = 23.34 °C - 25.00 °C = -1.66 °C Now, plug in the values into the formula: q = (75.0 g)(4.18 J/g·°C)(-1.66 °C) = -520.47 J The heat change is -520.47 J, which means the heat is released by the system (exothermic reaction) during the dissolution process.
02

Calculate the moles of NH4NO3 dissolved

We can find the moles of NH4NO3 dissolved (n) using the molar mass and given mass: Mass of NH4NO3: 1.60 g Molar mass of NH4NO3: N(14.01 g/mol) + H(4 x 1.01 g/mol) + N(14.01 g/mol) + O(3 x 16.00 g/mol) = 80.05 g/mol To find the moles of NH4NO3 dissolved, we can use the formula: n = (mass of NH4NO3) / (molar mass of NH4NO3) n = (1.60 g) / (80.05 g/mol) = 0.02 mol
03

Calculate the enthalpy change (ΔH) per mole of NH4NO3

Now that we have the heat change and the number of moles of NH4NO3 dissolved, we can calculate the enthalpy change (ΔH) in kJ/mol. To find the enthalpy change, we simply divide the heat change (q) by the moles of NH4NO3 (n), and convert J to kJ: ΔH = (q / n) / 1000 = (-520.47 J) / (0.02 mol) / 1000 = -26.02 kJ/mol The enthalpy change for the dissolution of NH4NO3 is -26.02 kJ/mol.

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Most popular questions from this chapter

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at 1.00 atm and $100 .^{\circ} \mathrm{C}\( occupies a volume of 30.6 \)\mathrm{L}$ . When 1 mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 1.00 atm and $100 .^{\circ} \mathrm{C}, 40.66 \mathrm{kJ}\( of heat is released. If the density of \)\mathrm{H}_{2} \mathrm{O}(l)$ at this temperature and pressure is \(0.996 \mathrm{g} / \mathrm{cm}^{3},\) calculate \(\Delta E\) for the condensation of 1 mole of water at 1.00 \(\mathrm{atm}\) and \(100 .^{\circ} \mathrm{C}\)

Explain the advantages and disadvantages of hydrogen as an alternative fuel.

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.

Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ} $$ a. How much heat is evolved for the production of 1.00 mole of $\mathrm{H}_{2} \mathrm{O}(l) ?$ b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when 186 \(\mathrm{g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was $2.0 \times 10^{8} \mathrm{L}\( at 1.0 atm and \)25^{\circ} \mathrm{C} .$ How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 $\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .$
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