ap chemistry 62 and 67 please 338 Chapter 13: Chemical Equilibrium At a certain

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ap chemistry 62 and 67 please

338 Chapter 13: Chemical Equilibrium At a certain temperature K for the reaction 2NO, ,O, is 7.5 liters/mole. If 2.0 moles of NO are ace mAL.0-liter container and permitted to react at this temperature, calculate the concentration at equilibrium. a) 0.39 moles/liter b) 0.65 moles/liter c) 0.82 moles/liter d) 7.5 moles/liter e) none of these Ans: a Algorithm: No Chapter/Section: 13.6 Difficuly: Difficult Keyword I: Chemistry Keyword 2: general chemistry Keyword 3: chemical equilibrium Keyword 4: using the equilibrium constant Keyword 5: calculating equilibrium concentrations 63. Exactly 1.0 mol N,O, is placed in an empty 1.0-1 container and is allowed to reach equilibrium desecmted by the equation N204(g) 2NO2(g) If at equilibrium the N204 is 32% dissociated, what is the value of the equilibrium constant, K , for the

Explanation / Answer

the reaction is 2NO2 (g)<---> N2O4(g)

2 moles of NO2 are placed in 2 L . concentration of NO2= 2 moles/2L=1M

let x= concentration of N2O4 at equilibroum, at equilibrium, concentration of NO2= 1-2x

K= equilibrium constant = [N2O4]/ [NO2]2= 7.5

x/(1-2x)2= 7.5, when solved using excel, x= 0.3865 moles/liter

concentration of N2O4 at equilibrium =0.3865 moles/L ( A is correct)

2. The reaction is CaCO3------->CaO+CO2, Kp= PCO2 , where P= pressure of CO2=1.16

P= 1.16 atm, V= 10L, T= 800 deg.c= 800+273= 1073K, R= gas constant =0.0821L.atm/mole.K

moles of CO2 formed at equilibrium, n= PV/RT=1.16*10/(0.0821*1073)= 0.132

molar mas of CaCO3= 100 g/mole, mass of CaCO3= 21.1 gm, moles of CaCO3= mass/molar mass

moles of CaCO3 taken = 21.1 gm/100=0.211 moles

from the reaction, 1 mole of CO2 is formed from 1 mole of CaCO3 decomposition.

hence moles of CaCO3 decomposed to form 0.132 moles of CO2= 0.132

% of CaCO3 reacted =100*0.132/0.211 = 62.56%( B is correct)

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