Complete and balance the reactions corresponding to the set with the last digit of your Student ID Number. Then find the standard enthalpy change for each reaction, using data from the Table of Thermodynamic Quantities.
Set 0)
Solid strontium chloride reacting with aqueous carbonate ion.
Aqueous nitric acid reacting with solid calcium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 1)
Solid magnesium chloride reacting with aqueous carbonate ion.
Aqueous nitric acid reacting with solid barium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 2)
Solid calcium chloride reacting with aqueous carbonate ion.
Aqueous nitric acid reacting with solid zinc(II) hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 3)
Solid barium chloride reacting with aqueous carbonate ion.
Aqueous nitric acid reacting with solid nickel(II) hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 4)
Solid iron(II) chloride reacting with aqueous carbonate ion.
Aqueous nitric acid reacting with solid magnesium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 5)
Solid silver nitrate reacting with aqueous sulfate ion.
Aqueous sulfuric acid reacting with solid sodium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 6)
Solid silver nitrate reacting with aqueous chloride ion.
Aqueous sulfuric acid reacting with solid potassium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 7)
Solid silver nitrate reacting with aqueous bromide ion.
Aqueous sulfuric acid reacting with aqueous sodium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 8)
Solid silver nitrate reacting with aqueous iodide ion.
Aqueous sulfuric acid reacting with aqueous potassium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 9)
Solid silver nitrate reacting with aqueous sulfide ion.
Aqueous sulfuric acid reacting with solid lithium hydroxide.
Aqueous bicarbonate ion reacting with aqueous cyanide ion.
Set 0)
SrCl2(s) + CO32–(aq) → SrCO3(s) + 2 Cl–(aq)
ΔH° = [(–1218) + 2(–167.2)] – [(–828.4) + (–677.1)] = –47 kJ/mole
2 HNO3(aq) + Ca(OH)2(s) → 2 H2O(l) + Ca2+(aq) + 2 NO3–(aq)
ΔH° = [2(–285.8) + (–542.96) + 2(–205.0)] – [2(–207.4) + (–986.1)] = –123.7 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO32–(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 1)
MgCl2(s) + CO32–(aq) → MgCO3(s) + 2 Cl–(aq)
ΔH° = [(–1096) + 2(–167.2)] – [(–641.3) + (–677.1)] = –112 kJ/mole
2 HNO3(aq) + Ba(OH)2(s) → 2 H2O(l) + Ba2+(aq) + 2 NO3–(aq)
ΔH° = [2(–285.8) + (–537.6) + 2(–205.0)] – [2(–207.4) + (–946.0)] = –158.4 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO32–(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 2)
CaCl2(s) + CO32–(aq) → CaCO3(s) + 2 Cl–(aq)
ΔH° = [(–1207) + 2(–167.2)] – [(–795.8) + (–677.1)] = –69 kJ/mole
2 HNO3(aq) + Zn(OH)2(s)→2 H2O(l) + Zn2+(aq) + 2 NO3–(aq)
ΔH° = [2(–285.8) + (–153.9) + 2(–205.0)] – [2(–207.4) + (–641.9)] = –78.8 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO32–(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 3)
BaCl2(s) + CO32–(aq) → BaCO3(s) + 2 Cl–(aq)
ΔH° = [(–1216) + 2(–167.2)] – [(–858.4) + (–677.1)] = –15 kJ/mole
2 HNO3(aq) + Ni(OH)2(s) → 2 H2O(l) + Ni2+(aq) + 2 NO3–(aq)
ΔH° = [2(–285.8) + (–64.0) + 2(–205.0)] – [2(–207.4) + (–529.7)] = –101.1 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO32–(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 4)
FeCl2(s) + CO32–(aq) → FeCO3(s) + 2 Cl–(aq)
ΔH° = [(–740.6) + 2(–167.2)] – [(–341.8) + (–677.1)] = –56.1 kJ/mole
2 HNO3(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg2+(aq) + 2 NO3–(aq)
ΔH° = [2(–285.8) + (–466.9) + 2(–205.0)] – [2(–207.4) + (–924.7)] = –109.0 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO3(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 5)
2 AgNO3(s) + SO42–(aq) → Ag2SO4(s) + 2 NO3–(aq)
ΔH° = [(–715.9) + 2(–205.0)] – [2(–124.4) + (–909.3)] = +32.2 kJ/mole
H2SO4(aq) + 2 NaOH(s) → 2 H2O(l) + 2 Na+(aq) + SO42–(aq)
ΔH° = [2(–285.8) + 2(–240.1) + (–909.3)] – [(–909.3) + 2(–425.6)] = –200.6 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO3(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 6)
AgNO3(s) + Cl–(aq) → AgCl(s) + NO3–(aq)
ΔH° = [(–127.1) + (–205.0)] – [(–124.4) + (–167.2)] = –40.5 kJ/mole
H2SO4(aq) + 2 KOH(s) → 2 H2O(l) + 2 K+(aq) + SO42–(aq)
ΔH° = [2(–285.8) + 2(–252.4) + (–909.3)] – [(–909.3) + 2(–424.8)] = –226.8 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO3(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 7)
AgNO3(s) + Br–(aq) → AgBr(s) + NO3–(aq)
ΔH° = [(–100.4) + (–205.0)] – [(–124.4) + (–120.9)] = –60.1 kJ/mole
H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + 2 Na+(aq) + SO42–(aq)
ΔH° = [2(–285.8) + 2(–240.1) + (–909.3)] – [(–909.3) + 2(–469.2)] = –113.4 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO3(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 8)
AgNO3(s) + I–(aq) → AgI(s) + NO3–(aq)
ΔH° = [(–61.84) + (–205.0)] – [(–124.4) + (–55.19)] = –87.3 kJ/mole
H2SO4(aq) + 2 KOH(aq) → 2 H2O(l) + 2 K+(aq) + SO42–(aq)
ΔH° = [2(–285.8) + 2(–252.4) + (–909.3)] – [(–909.3) + 2(–482.4)] = –111.6 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO3(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole
Set 9)
2 AgNO3(s) + S2–(aq) → Ag2S(s) + 2 NO3–(aq)
ΔH° = [(–31.8) + 2 (–205.0)] – [2(–124.4) + (41.8)] = –234.8 kJ/mole
H2SO4(aq) + 2 LiOH(s) → 2 H2O(l) + 2 Li+(aq) + SO42–(aq)
ΔH° = [2(–285.8) + 2(–278.5) + (–909.3)] – [(–909.3) + 2(–484.9)] = –158.8 kJ/mole
HCO3–(aq) + CN–(aq) → ← HCN(aq) + CO3(aq)
Kc ~ 10–11/10–10 ~ 0.1 < 100 so the reaction is an equilibrium.
ΔH° = [(105) + (–677.1)] – [(–691.11) + (151)] = –32 kJ/mole