뒤로General Chemistry Exam 2 Review: Solutions, Thermochemistry, and Kinetics
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Q1. What are two ways to increase the solubility of any gas in water?
Background
Topic: Gas solubility in solutions
This question tests your understanding of factors that affect the solubility of gases in liquids, specifically water.
Key Terms:
Solubility: The amount of a substance that can dissolve in a solvent at a given temperature and pressure.
Henry's Law: Relates the solubility of a gas to its partial pressure above the solution.
Step-by-Step Guidance
Recall that according to Henry's Law, the solubility of a gas increases as the partial pressure of the gas above the liquid increases.
Think about how temperature affects gas solubility: generally, increasing temperature decreases the solubility of gases in water.
Consider practical ways to manipulate these factors to increase solubility.
Try solving on your own before revealing the answer!
Final Answer:
Two ways to increase the solubility of a gas in water are:
Increase the pressure of the gas above the water (according to Henry's Law).
Decrease the temperature of the water (since gas solubility decreases with increasing temperature).
Both methods make it easier for gas molecules to dissolve in water.
Q2. The Henry’s Law constant for N2 (kH) is 1.67×10–4 M/atm. Suppose a diver is at a depth of 150 meters, where the pressure is 14.8 atm. What is the solubility of nitrogen, in moles/L, at this pressure? If a diver has 5.0 L of blood, how many grams of nitrogen are in the diver’s blood?
Background
Topic: Henry's Law and gas solubility
This question tests your ability to use Henry's Law to calculate the solubility of a gas at a given pressure, and then convert that solubility to mass for a given volume.
Key formula:
= solubility (mol/L)
= Henry's Law constant (mol/L·atm)
= pressure (atm)
Step-by-Step Guidance
Identify the values: M/atm, atm.
Use Henry's Law to calculate the solubility: .
Once you have the solubility in mol/L, multiply by the volume of blood (5.0 L) to find the total moles of nitrogen.
Convert moles of nitrogen to grams using the molar mass of N2 ( g/mol).
Set up the calculation for the final mass, but stop before computing the numeric value.
Try solving on your own before revealing the answer!
Final Answer:
Solubility: mol/L
Total moles in 5.0 L: mol
Mass of N2: g
There are approximately 0.346 grams of nitrogen dissolved in the diver's blood at this pressure.
Q3. Determine whether the following processes are exothermic or endothermic for the system. Explain. In each situation, the system of interest is underlined.
Background
Topic: Thermochemistry – heat flow and enthalpy
This question tests your understanding of exothermic and endothermic processes, and how to identify the direction of heat flow for a given system.
Key Terms:
Exothermic: Process where the system releases heat to the surroundings.
Endothermic: Process where the system absorbs heat from the surroundings.
System: The part of the universe being studied.
Surroundings: Everything outside the system.
Step-by-Step Guidance
For each scenario, identify the system and what is happening to it (e.g., freezing, melting, warming, mixing).
Recall that freezing and melting involve phase changes; consider whether heat is absorbed or released.
For mixing and temperature changes, think about whether the system feels cold or warm, and what that implies about heat flow.
Explain your reasoning for each process, but stop before stating the final classification for all cases.
Try solving on your own before revealing the answer!
Final Answer:
a) Freezing: Exothermic (system releases heat to surroundings)
b) Propane heater: Exothermic (system releases heat)
c) Melting beeswax: Endothermic (system absorbs heat)
d) Mixing NH4Cl: Endothermic (system absorbs heat, test tube feels cold)
Each answer is based on whether the system absorbs or releases heat.
Q4. A sample of iron having a mass of 1.03 kg required 254 kJ of heat to melt completely. What is the ΔHfus of iron? Fe = 55.85 g/mol.
Background
Topic: Enthalpy of fusion (ΔHfus)
This question tests your ability to calculate the enthalpy change associated with melting a substance, using mass and heat absorbed.
Key formula:
= heat absorbed (kJ)
= number of moles
Step-by-Step Guidance
Convert the mass of iron from kg to grams: .
Calculate the number of moles of iron: .
Use the formula , where kJ.
Set up the calculation for , but stop before computing the final value.
Try solving on your own before revealing the answer!
Final Answer:
Number of moles: mol
kJ/mol
The enthalpy of fusion for iron is approximately 13.8 kJ/mol.
Q5. 7.50 g of sodium acetate (CH3COONa) is added to 100.0 mL of water in a coffee cup calorimeter. The temperature of the solution changes from 20.00 °C to 23.52 °C. If the specific heat of the solution is 4.184 J/(g×°C), then what is the qreaction and ΔHreaction?
Background
Topic: Calorimetry and enthalpy change
This question tests your ability to use calorimetry data to calculate the heat of reaction and enthalpy change.
Key formulas:
= total mass of solution (g)
= specific heat (J/g·°C)
= change in temperature (°C)
= moles of sodium acetate
Step-by-Step Guidance
Calculate the total mass of the solution: add mass of sodium acetate to mass of water (assume density of water is 1.00 g/mL).
Find : °C.
Calculate using .
Calculate moles of sodium acetate: (molar mass of CH3COONa).
Set up the calculation for , but stop before computing the final value.
Try solving on your own before revealing the answer!
Final Answer:
Total mass: g
J = 1.59 kJ
Moles: mol
kJ/mol
The reaction released 1.59 kJ of heat, and the enthalpy change per mole is 17.4 kJ/mol.
Q6. The combustion reaction of ethanol and its ΔH are shown below. C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = –1367 kJ a. How much heat is released when 10.0 g of ethanol is combusted? b. How many mL of ethanol (density = 0.789 g/mL) would have to be combusted to release -2000. kJ of heat?
Background
Topic: Stoichiometry and enthalpy of combustion
This question tests your ability to relate mass and volume of a reactant to the heat released in a combustion reaction.
Key formulas:
Moles =
Volume =
Molar mass of ethanol (C2H6O): g/mol
Step-by-Step Guidance
For part (a): Calculate moles of ethanol in 10.0 g.
Multiply moles by to find total heat released.
For part (b): Set up a proportion to find the mass of ethanol needed to release 2000 kJ.
Convert mass to volume using the density.
Stop before computing the final numeric values.
Try solving on your own before revealing the answer!
Final Answer:
a) Moles: mol
Heat released: kJ
b) Mass needed: g
Volume: mL
To release 2000 kJ, combust about 85.5 mL of ethanol.
Q7. What is the difference between q and ΔH? Consider the reaction above. If you added more ethanol, what would happen to the values of q and ΔH?
Background
Topic: Heat (q) vs. enthalpy (ΔH)
This question tests your understanding of the distinction between heat and enthalpy, and how they relate to the amount of substance.
Key Terms:
q: Heat absorbed or released by the system.
ΔH: Enthalpy change per mole of reaction.
Step-by-Step Guidance
Recall that is the total heat exchanged, while is the heat per mole under constant pressure.
Think about how increasing the amount of reactant affects and .
Set up your explanation, but stop before stating the final answer.
Try solving on your own before revealing the answer!
Final Answer:
increases as more ethanol is combusted, because more heat is released overall. remains constant, as it is the enthalpy change per mole of reaction.
Adding more ethanol increases , but stays the same.
Q8. Use Hess’s Law to calculate the enthalpy change ΔH (in kJ) for the following reaction: NO2(g) + CO(g) → NO(g) + CO2(g) Given the following thermochemical equations: 2 NO(g) + O2(g) → 2 NO2(g) ΔΗ° = −113.1 kJ 2 CO(g) + O2(g) → 2 CO2(g) ΔΗ° = −566.0 kJ
Background
Topic: Hess's Law
This question tests your ability to use Hess's Law to combine thermochemical equations and calculate the enthalpy change for a new reaction.
Key formula:
Or, manipulate given equations to match the target reaction.
Step-by-Step Guidance
Write the target reaction and identify which given equations need to be reversed or scaled.
Reverse and/or scale the equations as needed to match the stoichiometry of the target reaction.
Add the manipulated equations together, canceling species as appropriate.
Set up the calculation for , but stop before computing the final value.
Try solving on your own before revealing the answer!
Final Answer:
Manipulate the equations:
Reverse the first equation and divide by 2: NO2(g) → NO(g) + 0.5 O2(g), ΔH = +56.55 kJ
Divide the second equation by 2: CO(g) + 0.5 O2(g) → CO2(g), ΔH = -283.0 kJ
Add together: NO2(g) + CO(g) → NO(g) + CO2(g), ΔH = +56.55 + (-283.0) = -226.45 kJ
The enthalpy change for the reaction is -226.45 kJ.
Q9. When magnesium metal is burned in a pure carbon dioxide atmosphere, carbon and magnesium oxide form. The balanced equation is 2 Mg(s) + CO2(g) → 2 MgO(s) + C(s) ΔHrxn ΔHf for CO2 (g) is –393.5 kJ/mole ΔHf for MgO(s) is –601.6 kJ/mole Use these enthalpies of formation (ΔHf) to calculate ΔHrxn. (Hint: What is the enthalpy of formation for an element in its standard state?) Is this reaction exothermic or endothermic?
Background
Topic: Enthalpy of reaction from enthalpies of formation
This question tests your ability to use enthalpies of formation to calculate the enthalpy change for a reaction.
Key formula:
ΔHf for elements in their standard state is 0.
Step-by-Step Guidance
Write the enthalpy of formation values for each reactant and product.
Multiply each ΔHf by the stoichiometric coefficient.
Sum the products and reactants, then subtract reactants from products.
Set up the calculation for ΔHrxn, but stop before computing the final value.
Try solving on your own before revealing the answer!
Final Answer:
Products: 2 MgO(s) (), C(s) (0)
Reactants: 2 Mg(s) (0), CO2(g) ()
kJ
The reaction is exothermic because ΔHrxn is negative.
Q10. Calculate enthalpy of combustion of ethane using the average bond enthalpy values below. 2CH3CH3(g) + 7O2(g) → 4CO2(g) + 6H2O(l) Bond energies (kJ/mol): C–H: 413 C–C: 347 O=O: 498 C=O: 799 O–H: 463 What does the ΔH value tell you about the stability of the reactant bonds compared to the products?
Background
Topic: Bond enthalpy and reaction enthalpy
This question tests your ability to use bond energies to estimate the enthalpy change for a reaction, and interpret the result.
Key formula:
Step-by-Step Guidance
Count the number of each type of bond broken in the reactants.
Count the number of each type of bond formed in the products.
Multiply the number of bonds by their bond energies to find total energy for bonds broken and formed.
Set up the calculation for ΔHrxn, but stop before computing the final value.
Think about what a negative ΔH means for bond stability.
Try solving on your own before revealing the answer!
Final Answer:
ΔHrxn = (energy of bonds broken) – (energy of bonds formed) = 13,858 – 18,370 = –4,512 kJ
A negative ΔH means the products are more stable (stronger bonds) than the reactants.
Q11. Define reaction rate. How does the rate of a reaction change as the reaction proceeds?
Background
Topic: Chemical kinetics
This question tests your understanding of reaction rate and how it changes over time.
Key Terms:
Reaction rate: The change in concentration of a reactant or product per unit time.
Step-by-Step Guidance
Recall the definition of reaction rate in terms of concentration and time.
Think about how reactant concentration changes as the reaction proceeds.
Consider how this affects the rate.
Try solving on your own before revealing the answer!
Final Answer:
Reaction rate is the change in concentration of a reactant or product per unit time, usually expressed as .
As the reaction proceeds, the rate typically decreases because reactant concentrations decrease.
Q12. In general, how does increasing the concentration of reactants change the rate of a reaction?
Background
Topic: Reaction rate and concentration
This question tests your understanding of how reactant concentration affects reaction rate.
Key Terms:
Rate law: Mathematical relationship between rate and concentration.
Step-by-Step Guidance
Recall that rate laws often show rate as proportional to reactant concentrations.
Think about how increasing concentration affects collision frequency.
Set up your explanation, but stop before stating the final answer.
Try solving on your own before revealing the answer!
Final Answer:
Increasing the concentration of reactants generally increases the rate of reaction, because more particles are available to collide and react.
Q13. Write four rate equations for the following reaction. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Background
Topic: Rate equations and stoichiometry
This question tests your ability to write rate expressions for each reactant and product based on their stoichiometric coefficients.
Key formula:
Step-by-Step Guidance
Identify the stoichiometric coefficients for each species.
Write the rate expression for each reactant (negative sign) and product (positive sign).
Set up the four rate equations, but stop before writing all the final expressions.
Try solving on your own before revealing the answer!
Final Answer:
Rate equations:
Q14. The following reaction: 2 MnO4– + 5 H2C2O4 + 6 H+ → 2 Mn2+ + 10CO2 + 8 H2O has the rate law: rate = k[MnO4–][H2C2O4][H+] a. What is the order with respect to each reactant? b. What is the overall order of the reaction? c. How would the rate change if the MnO4– concentration was doubled? d. How would the rate change if the H+ concentration was increased by a factor of four? e. How would the rate change if all reactant concentrations were doubled?
Background
Topic: Rate law and reaction order
This question tests your ability to interpret a rate law and predict how changes in concentration affect the rate.
Key formula:
Step-by-Step Guidance
Identify the exponent for each reactant in the rate law to determine the order with respect to each.
Add the exponents to find the overall order.
For each scenario, use the rate law to predict how the rate changes when concentrations are altered.
Set up the calculations for each part, but stop before stating the final answers.
Try solving on your own before revealing the answer!
Final Answer:
a) Order with respect to each reactant: 1
b) Overall order: 3
c) Doubling MnO4– doubles the rate.
d) Increasing H+ by four increases rate by four.
e) Doubling all reactants increases rate by times.
Q15. The A concentration in the reaction A → Products was monitored as a function of time. In the first 10 seconds, the reactant concentration decreased from 1.5 M to 0.75 M. It took another 10 seconds for the reactant concentration to drop from 0.75 M to 0.375 M. Is the reaction first or second order? Explain.
Background
Topic: Reaction order determination
This question tests your ability to use concentration vs. time data to determine the order of a reaction.
Key formulas:
First order:
Second order:
Step-by-Step Guidance
Notice that the concentration halves every 10 seconds.
Recall that for a first order reaction, the half-life is constant.
For a second order reaction, the half-life increases as concentration decreases.
Set up your explanation, but stop before stating the final answer.
Try solving on your own before revealing the answer!
Final Answer:
The reaction is first order, because the concentration halves every 10 seconds, indicating a constant half-life.
Q16. Calculate the average rate of reaction in the first 10 seconds of the reaction in the previous problem.
Background
Topic: Average reaction rate
This question tests your ability to calculate the average rate of a reaction using concentration and time data.
Key formula:
Step-by-Step Guidance
Identify the initial and final concentrations: M, M.
Calculate .
Divide by the time interval ( s).
Set up the calculation for average rate, but stop before computing the final value.
Try solving on your own before revealing the answer!
Final Answer:
M
Average rate: M/s
The average rate is 0.075 M/s in the first 10 seconds.