Science 10
1.
Brakke
ECA – Chem – Topic 01
Worksheet Thermochemistry II – Answer Key
Name ______KEY (11.26.10)______________________
When 1.34 grams of potassium bromide (KBr) dissolves in 74.0 grams of water in a coffee-cup calorimeter, the temperature
drops from 291.0 K to 290.3 K. Assume all heat absorbed in the solution process comes from the water.
a. Write a balanced equation for the solution process.
KBr (s) + H2O (l)  K+ (aq) + Br- (aq) + H2O
OR
KBr (s)  K+ (aq) + Br- (aq)
b. What is q of the water? Cannot find q of KBr directly so must find it for H2O and use q = -q
qH2O = mcΔT
qH2O = (74.0 g)(4.184 J/goc)(- 0.7oc) = - 216 J
mH2O = 74.0 g
cH2O = 4.184 J/goc
ΔTH2O = 290.3-291.0 = - 0.7 K = - 0.7oc
c. What is q when potassium bromide dissolves?
qBr = -(qH2O)
qBr = -(-216J)
qBr = +216J
d. Is the process endothermic or exothermic? Explain.
q for the salt is positive (+) and therefore energy is taken in from the surroundings (water) making the heat change for
KBr endothermic
e. What is the molar heat of solution if 1.34 grams of KBr were dissolved?
𝑞𝐾𝐵𝑟 =
+216 𝐽
1.34 𝑔 𝐾𝐵𝑟
𝑥
119𝑔 𝐾𝐵𝑟
1 𝑚𝑜𝑙 𝐾𝐵𝑟
= 19200
𝐽
𝑚𝑜𝑙
𝐾𝐵𝑟 = 19.2 𝐾𝐽/𝑚𝑜𝑙 = ΔH
f. Draw an energy diagram for this reaction
K+ (aq) + Br- (aq)
Enthalpy
ΔH = 19.2 𝐾𝐽/𝑚𝑜𝑙
Enthalpy
KBr (s)
g. Level 5:Calculate the energy change (∆H) when 10.00 moles of potassium bromide dissolves in
sufficient water.
10𝑚𝑜𝑙 𝐾𝐵𝑟 𝑥
19.2 𝑘𝐽
𝑚𝑜𝑙 𝐾𝐵𝑟
= ΔH = 192 kJ
h. Level 6:Calculate the energy change (∆H) when 10.00 grams of potassium bromide dissolves in
sufficient water.
10.00 𝑔𝐾𝐵𝑟 𝑥
1𝑚𝑜𝑙 𝐾𝐵𝑟
119 𝑔 𝐾𝐵𝑟
𝑥
19.2 𝑘𝐽
1 𝑚𝑜𝑙 𝐾𝐵𝑟
= ΔH = 1.61 kJ
i. Level 7: How many grams of potassium bromide were dissolved if the energy change (∆H) was 1.00 kilojoules?
1.00 𝑘𝐽 𝑥
1 𝑚𝑜𝑙 𝐾𝐵𝑟
19.2 𝑘𝐽
𝑥
119 𝑔 𝐾𝐵𝑟
1 𝑚𝑜𝑙 𝐾𝐵𝑟
= 6.20 𝑔 𝐾𝐵𝑟
Science 10
Brakke
ECA – Chem – Topic 01
2. When 1.80 grams of silver nitrate (AgNO3) is dissolved in 55.0 grams of water in a coffee-cup calorimeter the temperature of
the water goes from 20.5 degrees C to 21.3 degrees C. Assume all heat absorbed in the solution process comes from the
water.
a. Write a balanced equation for the solution process.
AgNO3 (s) + H2O (l) Ag+ (aq) + NO3- (aq) + H2O
OR
AgNO3 (s)  Ag+ (aq) + NO3-
(aq)
a. What is q of the water? Cannot find q of AgNO3directly so must find it for H2O and use q = -q
qH2O = mcΔT
qH2O = (55.0 g)( 4.184 J/goc)(0.8oc) = + 184 J
mH2O = 55.0 g
cH2O = 4.184 J/goc
ΔTH2O = 21.3-20.5 = + 0.8oc
b. What is q when silver nitrate dissolves?
qAgNO3 = -(qH2O)
qAgNO3r = -(184 J)
qAgNO3 = -184 J
c. Is the process endothermic or exothermic? Explain
q for the salt is negative (-) and therefore energy is released to the surroundings (water) making the heat change for
AgNO3 exothermic
d. What is the molar heat of solution when 1.80 grams of silver nitrate dissolves?
−184 𝐽
169.88𝑔 𝐴𝑔𝑁𝑂3
𝐽
𝑞𝐴𝑔𝑁𝑂3 =
𝑥
= −17400
𝐴𝑔𝑁𝑂3 = −17.4 𝐾𝐽/𝑚𝑜𝑙 = ΔH
1.80 𝑔 𝐴𝑔𝑁𝑂3
1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
𝑚𝑜𝑙
e. Draw an energy diagram for this reaction.
AgNO3 (s)
Enthalpy
ΔH = −17.4 𝐾𝐽/𝑚𝑜𝑙
Enthalpy
Ag+ (aq) + NO3- (aq)
f. Level 5: Calculate the energy change (∆H)when 11.55 moles of silver nitrate are dissolved in sufficient
Water:
−17.4 𝑘𝐽
11.55 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3 𝑥
= ΔH = -201 kJ
𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
f. Level 6: Calculate the energy change (∆H) when 4.52 grams of silver nitrate are dissolved in sufficient water:
4.52𝑔 𝐴𝑔𝑁𝑂3 𝑥
1𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
169.88 𝑔 𝐴𝑔𝑁𝑂3
𝑥
−17.4𝑘𝐽
1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
= ΔH = - 0.463 kJ
g. Level 7: How many grams of silver nitrate were dissolved if the energy change (∆H) was 315 kilojoules?
315 𝑘𝐽 𝑥 |
1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
−17.4 𝑘𝐽
|𝑥
169.88 𝑔 𝐴𝑔𝑁𝑂3
1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
= 3808 𝑔 𝐴𝑔𝑁𝑂3
Because it’s change in energy, the value for
matter (g) will obviously be positive.