# hw 3 - mj6727 hw03 McCord (53580) 1 This print-out should...

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Unformatted text preview: mj6727 hw03 McCord (53580) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. NOTE: If you see a number with an e in it followed by a signed number, you need to realize that the e means 10 to the power of what follows. For example.. 3 . 56 e- 07 = 3 . 56 10 07 Quest is still being tweaked to properly display scientific notation numbers. Some DO display correctly, while others display via the e notation. 001 10.0 points An equilibrium in which processes occur con- tinuously, with NO NET change, is called 1. homogeneous equilibrium. 2. heterogeneous equilibrium. 3. static equilibrium. 4. dynamic equilibrium. correct Explanation: For a system at dynamic equilibrium, al- though the concentrations of the components do not change, the processes continue to oc- cur in the foward and reverse directions at the same rate. 002 10.0 points Explain why equilibium constants are dimen- sionless. 1. The statement is not true. Equilibrium constants have units that involve some multi- ple of atmospheres or moles per liter. 2. They are dimensionless because the pres- sures or concentrations we put in are all for the substances in their standard states. 3. Every concentration or pressure that en- ters into K c or K p is really divided by the cor- responding concentration or pressure of the substance in its standard state. correct 4. They are dimensionless because concen- trations and pressures have no units. 5. They are not really dimensionless but we must treat them as such in order to be able to take ln K in the expression G =- RT ln K . Explanation: The amount of each component is in terms of activity, which is the measured amount (concentration, pressure) divided by the amount of that component in its standard state in that unit. The units in the numera- tor and denominator are identical and cancel out. 003 10.0 points Consider the reaction 2 HgO(s) 2 Hg( ) + O 2 (g) . What is the form of the equilibrium constant K c for the reaction? 1. K c = [Hg] 2 [O 2 ] [HgO] 2 2. None of the other answers is correct. 3. K c = [O 2 ] [HgO] 2 4. K c = [O 2 ] correct 5. K c = [Hg] 2 [O 2 ] Explanation: Solids and liquids are not included in the K expression. 004 10.0 points The expression for K c for the reaction at equi- librium is 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g) 1. [NH 3 ] 4 [O 2 ] 5 [NO] 4 [H 2 O] 6 2. [NH 3 ] 4 [O 2 ] 5 3. [NO] 4 [H 2 O ] 6 mj6727 hw03 McCord (53580) 2 4. [NO] 4 [H 2 O] 6 [NH 3 ] 4 [O 2 ] 5 correct Explanation: The equation must be written with the ap- propriate formula and correctly balanced. K c is the equilibrium constant for species in so- lution and equals the mathematical product of the concentrations of the chemical prod- ucts, divided by the mathematical product of the concentrations of the chemical reactants....
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## This note was uploaded on 03/23/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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hw 3 - mj6727 hw03 McCord (53580) 1 This print-out should...

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