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Chapter 8: Problem 12

When one mole of KOH is neutralized by sulfuric acid, \(q=-56 \mathrm{~kJ} .\) At \(22.8^{\circ} \mathrm{C}, 25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is neutralized by \(50.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) \(\mathrm{KOH}\) in a coffee-cup calorimeter. What is the final temperature of the solution? (Use the assumptions in Question 11.)

Short Answer

Expert verified
Answer: The final temperature of the solution is 27.27°C.

Step by step solution

01

Determine the moles of limiting reactant

To find out which reactant is limiting, examine the moles of each reactant: Moles of H₂SO₄ = Volume × Concentration = (25 mL) × (0.5 mol/L) = 0.0125 mol Moles of KOH = Volume × Concentration = (50 mL) × (0.5 mol/L) = 0.025 mol Since KOH reacts with H₂SO₄ in a 2:1 ratio, the amount of KOH required for 0.0125 mol of H₂SO₄ is 0.025 mol which is exactly the available amount of KOH. Therefore, KOH is the limiting reactant.
02

Calculate the heat change during the reaction

The heat change (q) per mole of KOH during the neutralization is given as -56 kJ. Since all 0.025 mol of KOH will be reacting, the total heat change produced by the reaction is: Total heat change, q = moles × heat change per mole = 0.025 mol × (-56 kJ/mol) = -1.4 kJ
03

Calculate the mass of the solution

The total volume of the solution is given by adding the volume of H₂SO₄ and KOH: Total volume = 25 mL + 50 mL = 75 mL Assuming 1 g/mL as the density of the solution (which is very close to the density of water), we can find the mass of the solution: Mass = Volume × Density = 75 mL × 1 g/mL = 75 g
04

Calculate temperature change

We are given that the calorimeter is well-insulated, so we can assume that all the heat released by the reaction is absorbed by the solution. To calculate the temperature change of the solution, we can use the formula: ΔT = q / (mass × C), where ΔT is the temperature change, q is the heat change, mass is the mass of the solution, and C is the specific heat of water (4.18 J/gC). Since the heat change is negative (exothermic reaction), the temperature will increase. Convert q to J by multiplying with 1000: -q = 1.4 kJ × 1000 J/kJ = 1400 J ΔT = 1400 J / (75 g × 4.18 J/gC) = 4.47 °C
05

Calculate final temperature

The initial temperature of the solution is given as 22.8°C. The final temperature of the solution can be calculated by adding the temperature change to the initial temperature: Final temperature = Initial temperature + ΔT = 22.8°C + 4.47°C = 27.27°C Therefore, the final temperature of the solution is 27.27°C.

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Most popular questions from this chapter

Write thermochemical equations for the decomposition of one mole of the following compounds into the elements in their stable states at \(25^{\circ} \mathrm{C}\) and 1 atm. (a) ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) (b) sodium fluoride \((s)\) (c) magnesium sulfate \((s)\) (d) ammonium nitrate (s)

A typical fat in the body is glyceryl trioleate, \(\mathrm{C}_{57} \mathrm{H}_{104} \mathrm{O}_{6}\). When it is metabolized in the body, it combines with oxygen to produce carbon dioxide, water, and \(3.022 \times 10^{4} \mathrm{~kJ}\) of heat per mole of fat. (a) Write a balanced thermochemical equation for the metabolism of fat. (b) How many kilojoules of energy must be evolved in the form of heat if you want to get rid of five pounds of this fat by combustion? (c) How many nutritional calories is this? (1 nutritional calorie = \(1 \times 10^{3}\) calories)

In photosynthesis, the following reaction takes place: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \mathrm{O}_{2}(g)+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \quad \Delta H=2801 \mathrm{~kJ}\) (a) Calculate \(\Delta H\) when one mole of \(\mathrm{CO}_{2}\) reacts. (b) How many kilojoules of energy are liberated when \(15.00 \mathrm{~g}\) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is burned in oxygen?

How many mL of water at \(10^{\circ} \mathrm{C}\) ( 2 significant figures) must be added to \(75 \mathrm{~mL}\) of water at \(35^{\circ} \mathrm{C}\) to obtain a final temperature of \(19^{\circ} \mathrm{C} ?\) (Make the same assumptions as in Question 9.)

To produce silicon, used in semiconductors, from sand \(\left(\mathrm{SiO}_{2}\right)\), a reaction is used that can be broken down into three steps: $$ \begin{aligned} \mathrm{SiO}_{2}(s)+2 \mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+2 \mathrm{CO}(g) & & \Delta H=689.9 \mathrm{~kJ} \\ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(g) & & \Delta H=-657.0 \mathrm{~kJ} \\ \mathrm{SiCl}_{4}(g)+2 \mathrm{Mg}(s) \longrightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s) & & \Delta H=-625.6 \mathrm{~kJ} \end{aligned} $$ (a) Write the thermochemical equation for the overall reaction for the formation of silicon from silicon dioxide; \(\mathrm{CO}\) and \(\mathrm{MgCl}_{2}\) are byproducts. (b) What is \(\Delta H\) for the formation of one mole of silicon? (c) Is the overall reaction exothermic?
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