CHEM 120 Homework. Chapter 8.
a) Fist law of thermodynamics b) Second law of thermodynamics c) Law of conservation of mass
d) Law of mass action
1. Br2 (l) --> Br2 (g) 2. 2NO(g) + O2 (g) --> 2NO2 (g)
3. MgSO4 (s) --> MgO(s) + SO3 (g) 4. 2HgO(s) --> 2Hg(l) + O2 (g)
a. 1 and 2 b. 2 and 3 c. 3 and 4 d. 1, 3, and 4 e. 1, 2, 3, and 4
d) negative entropy of the system
a) enthalpy, H b) entropy, S c) free energy, G (DG =DH-TDS) d) temperature, T
a) 982 cal b) 4.75 cal c) 6.78 cal d) 2.57 cal e) 5.01 cal
a) thermodynamics of a reaction b) equilibrium position of a reaction
c) reaction rates d) calculating the amount of precuts in a reaction
a) reaction mechanism b) reaction energy diagram c) activation energy
d) activated complex
a) increase concentrations of reactants b)increase temperature c) add a catalyst d) increase the products
a) diffusion of the odor from cooking food b) dissolving of sugar in coffee
c) decomposition of iron ore into pure iron d) burning a candle
e) melting ice at 25 degrees Celsius
a) the tiny dust particles will result in a very high surface area.
b) the dust particles have a very flammable material.
c) flammable gases are always present in the elevators.
d) of the high temperatures always present in the elevators.
e) of the constant sources of ignition.
C3H8(g) + 5O2(g) <==> 3CO2(g) + 4H2O(g).
a) reactants are completely changed to products. b) there are equal amounts of reactants and products.
c) the rate at which reactants form products becomes zero.
d) the rate at which reactants form products is the same as the rate at which products form reactants.
PCl5(g) <==> PCl3(g) + Cl2(g); .H = +500 kJ
Which of the following changes will shift the equilibrium to the LEFT?
a) increasing temperature b) increasing volume c) increasing pressure d) removing Cl2(g)
e) adding PCl5(g)
COCl2(g) <===> CO(g) + Cl2(g)
At equilibrium, [CO] = 4.14 × 10-6 M; [Cl2] = 4.14 × 10-6 M; and [COCl2] = 0.0627 M. Calculate the value of the equilibrium constant.
a) 1.32 × 10-4 b) 1.51 × 104 c) 2.73 × 10-10 d) 3.66 × 109 e) 6.60 × 10-5