Chem 1151 Quiz 5-green 15 points April 8, 2008
Name______KEY______Sec______
1.(2) The First Law of Thermodynamics states
A. H = E +PV
B. w = -PΔV
C. q = sm ΔT
D. ***ΔE = q + w
E. ΔH = qP
2.(2) The reaction for ΔH0f for Ca(NO3)2(s) is
A. Ca(s) + 2N(s) + 6O(g) à Ca(NO3)2(s)
B. ***Ca(s) + N2(s) + 3O2(g) à Ca(NO3)2(s)
C. Ca(NO3)2(s) à Ca(s) + N2(s) + 3O2(g)
D. 2Ca(s) + 2N2(s) + 6O2(g) à 2Ca(NO3)2(s)
E. Ca2+(aq) + 2NO3-(aq) à Ca(NO3)2(s)
3.(4) Select the correct sign for the property listed for CO2(g) à CO2(s)
a. q –
b. w +
4.(7) When calcium chloride dissolves in water,
CaCl2(s) à Ca2+(aq) + 2Cl-(aq) ΔH = -81.5 kJ
a. (2) What is ΔH for 2Ca2+(aq) + 4Cl-(aq) à 2CaCl2(s)?
(-2) (-81.5) = + 163 kJ
b. (2) How much heat is given off when 5.55 g of CaCl2 dissolves?
g CaCl2 à mol CaCl2 à q
5.55g/(111g/mol) (-81.5 kJ) = - 4.08 kJ or 4.08 kJ given off
c. (3) If the 5.55 g of CaCl2 dissolves in 600 mL of water, initially at 25 oC, how much does the temperature of the solution change? The specific heat of water = 4.18 J/(g- oC). Assume that the reaction takes place in an insulated container.
The heat given off in part b warms the soln in part c.
q = m s ΔT or ΔT = q/ms = 4080J/(600 g) * 4.18) = 1.63 oC
or if you use m = 605.55 g, then ΔT = 1.62 oC
Chem 1151 Quiz 5-blue 15 points April 8, 2008
Name______KEY______Sec______
1.(2) The First Law of Thermodynamics states
A. ***ΔE = q + w
B. H = E +PV
C. w = -PΔV
D. q = sm ΔT
E. ΔH = qP
2.(2) The reaction for ΔH0f for Mg(NO3)2(s) is
A. Mg(s) + 2N(s) + 6O(g) à Mg(NO3)2(s)
B. Mg(NO3)2(s) à Mg(s) + N2(s) + 3O2(g)
C. 2Mg(s) + 2N2(s) + 6O2(g) à 2Mg(NO3)2(s)
D. ***Mg(s) + N2(s) + 3O2(g) à Mg(NO3)2(s)
E. Mg2+(aq) + 2NO3-(aq) à Mg(NO3)2(s)
3.(4) Select the correct sign for the property listed for CO2(g) à CO2(s)
a. w +
b. q –
4.(7) When calcium chloride dissolves in water,
CaCl2(s) à Ca2+(aq) + 2Cl-(aq) ΔH = -81.5 kJ
a. (2) What is ΔH for 2Ca2+(aq) + 4Cl-(aq) à 2CaCl2(s)?
(-2) (-81.5) = + 163 kJ
b. (2) How much heat is given off when 16.7 g of CaCl2 dissolves?
g CaCl2 à mol CaCl2 à q
16.7g/(111g/mol) (-81.5 kJ) = - 12.3 kJ or 12.3 kJ given off
c. (3) If the 16.7 g of CaCl2 dissolves in 900 mL of water, initially at 25 oC, how much does the temperature of the solution change? The specific heat of water = 4.18 J/(g- oC). Assume that the reaction takes place in an insulated container.
The heat given off in part b warms the soln in part c.
q = m s ΔT or ΔT = q/ms = 12300J/(900 g * 4.18) = 3.27 oC
or if you use m = 916.7 g, then ΔT = 3.21 oC