00lmixturecontains0145molcof2 0262mo lofco2 and0074mo

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Unformatted text preview: l each of SO2 and NO2 are mixed in a 1.00­L flask and allowed to reach equilibrium. What percent of SO2 is converted to product? ˆˆ SO2 (g) + NO2 (g) ‡ˆ† SO3 (g) + NO (g) Initial D Equilibrium 1.50 M – x 1.50 –x K C = 1.50 M – x 1.50 –x 2 [SO 3 ][NO] x = = 3.00 2 [SO 2 ][NO 2 ] (1.50 ­ x) Taking square root of each side, x = 1.73 1.50 ­ x x + 1.73 x = 2.595 x = 0.951 % = 0 + x x 0. 951 x100 = 63.4% 1.50 3 0 + x x 8. The fo llowing equilibrium exists at 1000 ºC with KC = 2.00. ˆˆ 2 COF2 (g) ‡ˆ† CO2 (g) + CF4 (g) If a 5.00­L mixture contains 0.145 mol COF2 , 0.262 mo l of CO2 and 0.074 mo l of CF4 at 1000 ºC, in which direct ion will the mixture proceed to reach equilibrium? 0.145 mol 0.262 mol = 0.0290 M [CO 2 ] = = 0.0524 M 5.00 L 5.00 L 0.074 mol [CF4 ] = = 0.0148 M 5.00 L [CO 2 ][CF4 ] (0.0524)(0.0148) Q c = = = 0.922 2 [COF2 ]2 (0.0290) [COF2 ] = Since Q c < K c , reaction w ill proceed in the forw ard direction 9. Formamide deco mposes at high temperatures according to the equation shown below: ˆˆ HCONH2 (g) ‡ˆ† NH3 (g) + CO (g) KC = 4.84 at 400 K If 0.186 mo l of formamide is placed in a 2.16­L flask and allowed to decompose at 400 K, what will be the total pressure at equilibrium? PHCONH 2 = nRT (0.186)(0.0821)(400 K ) = = 2.83 atm V 2.16 L Kp = Kc (RT)Dn = 4.84 [(0.082...
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This document was uploaded on 03/18/2014 for the course CHEM 102 at Los Angeles Mission College.

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