Chapter 15 Solutions

# Chapter 15 Solutions - Chapter 15 Answers 7. Remember that...

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Chapter 15 Answers 7. Remember that K p =K c (RT) Δn . Since we are looking for K C , rearrange the equation to give K C =K p /(RT) Δn . Remember that Δn=(sum of the moles of gas of the products – sum of the moles of gas of the reactants) a) Δn=1 T=303K R=0.08206 L*atm*mol -1 K -1 Kp=2.9x10 -2 Plug it all into the equation: K C =K p /(RT) Δn =2.9x10 -2 /((0.08206x303) 1 ) =1.2x10 -3 Remember that Kc is unitless – we are technically multiplying by an invisible constant that conveniently cancels all the units. b) Δn=-1 T= 184°C = 457K Kp=1.48x10 4 K C =K p /(RT) Δn =1.48x10 4 /((0.08206x457) -1 ) =5.55x10 5 c) ) Δn=0, so Kp=Kc Kc=0.429 9. Vaporization of water is the process by which liquid water turns to gaseous water, so the equation of interest is H 2 O (l) H 2 O (g) Convert pressure to atm: 23.8Torr(1atm/760torr)= 0.0313atm Find Kp: Kp=P ° (P H2O(g) ) =0.0313 For Kc, it is first necessary to determine Δn. Since we have only one gas (in the products), Δn=1 K C =K p /(RT) Δn Kc=0.0313/((0.0313x298K) 1 ) =1.28x10 -3

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13. For this question it is important to note that “when individual equations are added, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction” (p. 630 of your textbook). Taking that into account: Equation A: C graphite + CO 2(g) 2CO (g) (Kc=0.64) Equation B: CO 2(g) + H 2(g) CO (g) + H 2 O (g) (Kc=1.4) Equation C: C graphite + ½ O 2(g) CO (g) (Kc=1x10 8 ) To get our overall equation of: 2 H 2(g) + O 2(g) 2 H 2 O (g) we need Equation A x (-2) Equation B x (2) Equation C x (2) Kc = (0.64) -2 (1.4) 2 (1.8x10 8 ) 2 =5x10 16 Next is determining Kp. For that we can go back to the trusty equation K p =K c (RT) Δn . For the overall equation, Δn=-1 K p =K c (RT) Δn =5x10 16 (0.08206x1200) -1 =5x10 14 21. a) The chemical equation (balanced) is: 4NH 3(g) + 7O 2(g) 4NO 2(g) + 6H 2 O(g) This has to be adjusted to reflect that the Kc for NH 3 oxidation is on a ‘per mole’ basis – so we divide all the coefficients by 4: NH 3(g) + 7/4 O 2(g) NO 2(g) + 3/2 H 2 O(g) b) We are given equations with which to solve for Kp: 1x: NH 3(g) + 5/4 O 2(g) NO (g) + 3/2 H 2 O(g) (Kp=2.11x10 19 ) -1x: NO 2(g) NO (g) + ½ O 2(g) (Kp=0.524) Now we can get the Kp for the overall equation: Kp=(2.11x10 19 )(0.524) -1 =4.03x10 19
23. a) To find Kc, you need to have the concentrations of the reactants and products. Concentration is measured in moles per liter (or moles per unit volume).

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## This note was uploaded on 04/04/2011 for the course SCIENCE CHEM 120 taught by Professor Fenster during the Winter '11 term at McGill.

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Chapter 15 Solutions - Chapter 15 Answers 7. Remember that...

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