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The answers and explanations for a general chemistry exam held in spring 2011. It covers topics such as entropy, enthalpy, spontaneity, ideal gas law, and thermodynamic justifications for spiritual practices.
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This is a sample test he gave us to take before the first exam. He originally highlighted all of the correct answers but that defeats the purpose of taking the test, so I put all of the correct answers at the bottom of the test instead. Good luck! Exam 2 – ANSWER KEY TOTAL POINTS = 98 General Chemistry Spring 2011
each molecule of gas. d) The pressure of a gas in a closed container will increase if the container volume is decreased.
-5130 kJ/mol lactose x 1 mol lactose/ 12 mol O2 = - 427.5 kJ/mol O
more heat released on formation of a N-N triple bond than the heat required to break 4 N-H bonds and a O-O bond. There is in addition, heat produced by the formation of additional O-H bonds.
liquid N2 at –200 deg C. There is no change in pressure—atmospheric pressure on the balloon remains the same— and there is no change n the number of moles of gas. So V (of balloon) decreases.
reaction is stated to be spontaneous, so that deltaG must be < 0. Since delta G = deltaH – TdeltaS, the only way for deltaG to be negative is if deltaH is very negative to compensate for the –TdeltaS term that will be positive.
of respiration is negative and spontaneous) to keep them alive through the oxidation of carbohydrates, like glucose. The carbohydrates originate from plants, whch make the carbohydrates through the process of photosynthesis. Photosynthesis is not a spontaneous process (free energy is positive) but is made possible through the energy in the form of light from the sun. The sun is responsible for all life on earth.
First balance the equation! 2Fe3O4 + 1/2 O2 --> 3Fe2O delta G(rxn) = delta G(products) - delta G(rgts) = 3 delta G (Fe2O3) – 2 delta G (Fe3O4) – 1/ delta G (O2) = = 3 (-742.2) – 2(-1015.4) – 1/2 (O) = -195.8 kJ (for formation of 3 mole Fe2O3 ) If the reaction is balanced as: 4Fe3O4 + O2 ‡ 6Fe2O delta G(rxn) = delta G(products) - delta G(rgts) = 6 delta G (Fe2O3) – 4 delta G (Fe3O4) – delta G (O2) = = 6(-742.2) – 4(-1015.4) – 1/2 (O) = -391.6 kJ (for formation of 6 mole Fe2O3)
the elements in their most stable state at 25 deg C and 1 bar pressure.
spontaneous.
C2H5OH (l) --> C2H5OH (g) Therefore for delta G = deltaH – TdeltaS, deltaG = 0. And delta H = Tdelta S Find deltaH and deltaS values for C2H5OH (l) and C2H5OH (g). delta H ( C2H5OH , (g)) - delta H ( C2H5OH , (l )) = T delta S ( C2H5OH , (g)) - delta S ( C2H5OH , (l)) -235.1 – (-277.69) = T (282.7 – 160.7) 42.59 kJ/mol = T(0.122 kJ/K mol) T = 349.1 K = 76.1 deg C.
deltaH = mass x heat capacity x change temperature so: delta H (spoon) = - deltaH (water+glass) 100g x 0.22 J/ g deg x (T(final) – 93 deg C) = -(250 g x 4.18 J/ g deg x (T(final) – 25 deg C) T(final) = 26.4 deg C.