Download electrical energy and capacitance

MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 6: CHEMICAL REACTIONS (PART 2)
WORKSHEET SOLUTIONS
1. Which of the following increases the rate of reaction?
(A) increasing the product concentration
(B) decreasing the product concentration
(C) decreasing the temperature
(D) decreasing the pressure
(E) decreasing the surface area of the reactants
2. Increasing the surface area of the reactants has what effect on chemical reactions?
(A) increases pressure
(B) decreases reaction rate
(C) decreases pressure
(D) increases reaction rate
(E) no effect
Questions 3 - 6. Classify each reaction as: gas evolution, combustion, neutralization, or redox.
3. Ag(s) + Al+3(aq)  Al(s) + 3 Ag+(aq)
3A.
redox
4. HI(aq) + KOH(aq)  H2O(l) + KI(aq)
4A.
neutralization
5. H2SO3(aq)  SO2(g) + H2O(l)
5A.
gas evolution
6. CH4(g) + 2 O2(g)  CO2(g) + H2O(l)
6A.
combustion
Questions 7 - 10. Determine (i) voltage and (ii) reaction/no reaction for each of the following redox
reactions.
7. F2(g) + Ni(s) 
7A.
(1) Determine the half-reactions:
(2) F2(g) + 2 e-  2 F(3) Ni(s)  Ni+2 + 2 e(4) Determine the voltages:
(5) F2(g) + 2 e-  2 F- = 2.87 V
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7A. (continued…)
(6) Ni(s)  Ni+2 + 2 e- = 0.25 V (reverse reaction)
(7) 2.87 V + 0.25 V
(8) 3.12 V. Possible reaction.
8. Cu(s) + Mg+2 
8A.
(1) Determine the half-reactions:
(2) Cu(s)  Cu+2 + 2 e(3) Mg+2 + 2 e-  Mg(s)
(4) Determine the voltages:
(5) Cu(s)  Cu+2 + 2 e- = - 0.42 V (reverse reaction)
(6) Mg+2 + 2 e-  Mg(s) = - 2.37 V
(7) - 0.42 V – 2.37 V
(8) - 2.79 V. No reaction.
9. Ca(s) + Mg+2 
9A.
(1) Determine the half-reactions:
(2) Ca(s)  Ca+2 + 2 e(3) Mg+2 + 2 e-  Mg(s)
(4) Determine the voltages:
(5) Ca(s)  Ca+2 + 2 e- = 2.87 V (reverse reaction)
(6) Mg+2 + 2 e-  Mg(s) = - 2.37 V
(7) 2.87 V – 2.37 V
(8) 0.50 V. Possible reaction.
10. Cl2(g) + Co(s) 
10A.
(1) Determine the half-reactions:
(2) Cl2(g) + 2 e-  2 Cl(3) Co(s)  Co+2 + 2e(4) Determine the voltages:
(5) Cl2(g) + 2 e-  2 Cl- = 1.36 V
(6) Co(s)  Co+2 + 2 e- = 0.28 V (reverse reaction)
(7) 1.36 V + 0.28 V
(8) 1.64 V. Possible reaction.
Questions 11 - 13. Balance the following reactions.
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
11. Ethanol (C2H6O) has been considered as a fuel for automobiles. Complete a balanced combustion
reaction for ethanol.
11A.
(1) Set up a skeletal combustion equation:
__ C2H6O(l) + __ O2(g)  __ CO2(g) + __ H2O(l)
(2) Solve the skeletal equation:
C2H6O(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
12. HC2H3O2(aq) + NaHSO3(aq) 
12A.
(1) Identify the reactant ions:
(2) HC2H3O2 = H+ + C2H3O2(3) NaHSO3 = Na+ + HSO3(4) Swap opposite charges:
(5) (H+ + HSO3-) + (Na+ + C2H3O2-)
(6) Identify the possible products: H2SO3 + NaC2H3O2
(7) Set up a skeletal equation:
HC2H3O2 + NaHSO3  H2SO3 + NaC2H3O2
(8) Notice how H2SO3 (sulfurous acid) is a reaction intermediate:
H2SO3  H2O(l) + SO2(g)
(9) Substitute the decomposition products back into the skeletal equation:
HC2H3O2 + NaHSO3  H2O + SO2 + NaC2H3O2
(10) Final balanced equation:
HC2H3O2(aq) + NaHSO3(aq)  H2O(l) + SO2(g) + NaC2H3O2(aq)
13. HCl(aq) + Ba(OH)2(aq) 
13A.
(1) Identify the reactant ions:
(2) HCl = H+ + Cl(3) Ba(OH)2 = Ba+2 + 2 OH(4) Swap opposite ions: (H+ + OH-) + (Ba+2 + Cl-)
(5) Determine the products: H2O + BaCl2
(6) Write a skeletal equation:
__ HCl + __ Ba(OH)2  __ BaCl2 + __ H2O
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
13A. (continued…)
(7) Determine which products are soluble.
(8) According to the solubility table: BaCl2 is soluble.
(9) Final balanced equation:
2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2 H2O(l)
14. Water is added to the following reaction:
H2O(l)   OH-(aq) + H+(aq)
What additional substances are created?
14A.
Adding more reactant increases the amount of product, so OH- and H+ increase.
15. Pressure is added to the following reaction:
H2O(l)   OH-(aq) + H+(aq)
What additional substances are created?
15A.
This equation has one mole of reactants and two moles of products in equilibrium. According to Le
Chatelier, increased pressure drives the production of more H2O.
16. The following reaction is exothermic in the forward direction:
2 CO(g) + O2(g)   2 CO2(g)
How would an increase in temperature affect both these substances?
16A. Re-write the expression as:
2 CO(g) + O2(g)   2 CO2(g) + Heat
According to Le Chatelier, raising the temperature results in less CO2 and more CO and O2.
17. Ba(NO3)2(aq) reacts with Na2CO3(aq). Answer the following questions. You do not need to
balance the equations.
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
17a. Write the molecular equation.
A.
(1) Identify the ions of the reactants:
(2) Ba(NO3)2 = Ba+2 + 2 NO3(3) Na2CO3 = 2 Na+ + CO3-2
(4) Swap opposite charges: (Ba+2 + CO3-2) + (Na+ + NO3-)
(5) Determine the possible products: BaCO3 + NaNO3
(6) Write a skeletal equation:
_Ba(NO3)2 + _Na2CO3  _BaCO3 + _NaNO3
(7) Determine which products are soluble or insoluble.
(8) According to the solubility table:
BaCO3 is insoluble.
NaNO3 is soluble.
(9) Final unbalanced molecular equation:
Ba(NO3)2(aq) + Na2CO3(aq)  BaCO3(s) + NaNO3(aq)
17b. Write the complete ionic equation.
A.
Ba+2(aq) + NO3-(aq) + Na+(aq) + CO3-2(aq)  BaCO3(s) + Na+(aq) + NO3-(aq)
17c. Write the net ionic equation.
A.
Ba+2(aq) + NO3-(aq) + Na+(aq) + CO3-2(aq)  BaCO3(s) + Na+(aq) + NO3-(aq)
Ba+2(aq) + CO3-2(aq)  BaCO3(s)
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 6: CHEMICAL REACTIONS (PART 2)
QUIZ SOLUTIONS
1. Which of the following increases the rate of reaction?
(A) increasing the product concentration
(B) decreasing the product concentration
(C) decreasing the temperature
(D) decreasing the pressure
(E) decreasing the surface area of the reactants
2. Increasing the surface area of the reactants has what effect on chemical reactions?
(A) increases pressure
(B) decreases reaction rate
(C) decreases pressure
(D) increases reaction rate
(E) no effect
3. OH- is added to the following reaction:
H2O(l)   OH-(aq) + H+(aq)
What additional substances are created?
3A.
Adding more product increases the amount of reactant, so H2O increases.
4. Pressure is added to the following reaction:
PCl5(g)   PCl3(g) + Cl2(g)
What additional substances are created?
4A.
This equation has one mole of reactants and two moles of products in equilibrium. According to Le
Chatelier, increased pressure drives the production of more PCl5.
5. The following reaction is exothermic in the forward direction:
N2(g) + O2(g)   2 NO(g)
How would an increase in temperature affect both these substances?
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
5A. Re-write the expression as:
N2(g) + O2(g)   2 NO(g) + Heat
According to Le Chatelier, raising the temperature results in less NO and more N2 and O2.
6. Ag(s) + Fe+3(aq)  Fe(s) + 3 Ag+(aq)
6A.
redox
7. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
7A.
neutralization
8. H2SO3(aq)  SO2(g) + H2O(l)
8A.
gas evolution
9. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(l)
9A.
combustion
10. Cl2(g) + Ni(s) 
10A.
(1) Determine the half-reactions:
(2) Cl2(g) + 2 e-  2 Cl(3) Ni(s)  Ni+2 + 2 e(4) Determine the voltages:
(5) Cl2(g) + 2 e-  2 Cl- = 1.36 V
(6) Ni(s)  Ni+2 + 2 e-= 0.25 V (this is a reverse reaction)
(7) 1.36 V + 0.25 V
(8) 1.61 volts.
11. Cu(s) + Zn+2 
11A.
(1) Determine the half-reactions:
(2) Cu(s)  Cu+2 + 2 e(3) Zn+2 + 2 e-  Zn(s)
(4) Determine the voltages:
(5) Cu(s)  Cu+2 + 2 e- = - 0.42 V (this is a reverse reaction)
(6) Zn+2 + 2 e-  Zn(s) = - 0.76 V
(7) - 0.42 V – 0.76 V
(8) - 1.18 volts.
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
12. Na(s) + Li+ 
12A.
(1) Determine the half-reactions:
(2) Na(s)  Na+ + e(3) Li+ + e-  Li(s)
(4) Determine the voltages:
(5) Na(s)  Na+ + e- = 2.71 V (reverse reaction)
(6) Li+ + e-  Li(s) = - 3.03 V
(7) 2.71 V - 3.03 V
(8) - 0.32 volts.
13. Mg(s) + Co+2 
13A.
(1) Determine the half-reactions:
(2) Mg(s)  Mg+2 + 2 e(3) Co+2 + 2 e-  Co(s)
(4) Determine the voltages:
(5) Mg(s)  Mg+2 + 2 e- = 2.37 V (reverse reaction)
(6) Co+2 + 2 e-  Co(s) = - 0.28 V
(7) 2.37 V - 0.28 V
(8) 2.09 volts.
14. Complete a balanced combustion reaction for C8H16O.
14A.
(1) Set up a skeletal combustion equation:
__ C8H15OH(l) + __ O2(g)  __ CO2(g) + __ H2O(l)
(2) Solve the skeletal equation:
2 C8H15OH(l) + 23 O2(g)  16 CO2(g) + 16 H2O(l)
15. NH4NO2(aq) + NaOH (aq) 
15A.
(1) Identify the reactant ions:
(2) NH4NO2 = NH4+ + NO2(3) NaOH = Na+ + OH(4) Swap opposite charges:
(5) (NH4+ + OH-) + (Na+ + NO2-)
(6) Identify the possible products:
NH4OH + NaNO2
(7) Set up a skeletal equation:
__ NH4NO2 + __ NaOH  NH4OH + __ NaNO2
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
15A. (continued…)
(8) Notice how NH4OH (ammonium hydroxide) is a reaction intermediate:
NH4OH  NH3(g) + H2O(l)
(9) Substitute the decomposition products back into the skeletal equation:
__ NH4NO2 + __ NaOH  NH3 + H2O + __ NaNO2
(10) Final balanced equation:
NH4NO2(aq) + NaOH(aq)  NH3(g) + H2O(l) + NaNO2(aq)
16. HClO3(aq) + KOH(aq) 
16A.
(1) Identify the reactant ions:
(2) HClO3 = H+ + ClO3(3) KOH = K+ + OH(4) Swap opposite ions: (H+ + OH-) + (K+ + ClO3-)
(5) Determine the products: H2O + KClO3
(6) Write a skeletal equation:
__ HClO3 + __ KOH  __ KClO3 + __ H2O
(7) Final balanced equation:
HClO3(aq) + KOH(aq)  KClO3(aq) + H2O(l)
17. Be(NO3)2(aq) reacts with K2CO3(aq). Answer the following questions. You do not need to
balance the equations.
17a. Write the molecular equation.
A.
(1) Identify the ions of the reactants:
(2) Be(NO3)2 = Be+2 + NO3(3) K2CO3 = K+ + CO3-2
(4) Swap opposite charges:
(Be+2 + CO3-2) + (K+ + NO3-)
(5) Determine the products:
BeCO3 + KNO3
(6) Write a skeletal equation:
_ Be(NO3)2 + _ K2CO3  _BeCO3 + _KNO3
(7) Determine which products are soluble or insoluble.
(8) According to the solubility table:
BeCO3 is insoluble.
KNO3 is soluble.
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
17a. (continued…)
(9) Final unbalanced molecular equation:
Be(NO3)2(aq) + K2CO3(aq)  BeCO3(s) + KNO3(aq)
17b. Write the complete ionic equation.
A.
Be+2(aq) + NO3-(aq) + K+(aq) + CO3-2(aq)  BeCO3(s) + K+(aq) + NO3-(aq)
17c. Write the net ionic equation.
A.
Be+2(aq) + NO3-(aq) + K+(aq) + CO3-2(aq)  BeCO3(s) + K+(aq) + NO3-(aq)
Be+2(aq) + CO3-2(aq)  BeCO3(s)
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