Unit7 Chemical equilibrium qns

Unit7 Chemical equilibrium qns - 1 QUEST TUTORIALS Head...

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Unformatted text preview: 1 QUEST TUTORIALS Head Office : E-16/289, Sector-8, Rohini, New Delhi, Ph. 65395439 UNIT 7 CHEMICAL EQUILIBRIUM Select the lettered item which best completes or answers the following : 1. Which one of the following is an illustration of a reversible reaction ? (a) Pb(NO 3 ) 2 + 2NaI PbCl 2 + 2NaNO 3 (b) AgNO 3 + NaCl AgCl + NaNO 3 (c) 2Na + 2H 2 O 2NaOH + H 2 (d) KNO 3 + NaCl KCl + NaNO 3 2. A reaction would go to an end if one of the products is (a) Al(NO 3 ) 3 (b) (NH 4 ) 2 S (c) CuSO 4 (d) FeCO 3 3. When a reversible reaction is in equilibrium, opposing forces (a) stop acting (b) are shifted to the right (c) are in constant operation (d) go to one end 4. The quantitative relationship between rate of reaction and concentration was established by (a) Guldburge and Waage (b) Le-Chatelier (c) Vant Hoff (d) Arrhenius 5. The value of K e of a reaction is affected by (a) T + P only (b) T and Catalyst only (c) P and Catalyst only (d) T only 6. At equilibrium, the free energy for the reaction is (a) zero (b) always negative (c) maximum (d) infinite 7. The standard free energy change ( G is related to equilibrium constant K e (K e or K P ) by the relation (a) G = 2.303 RT log K eq (b) G = RT log e K eq (c) log K eq = RT 303 . 2 G r - (d) all of these 8. For the reaction 2SO 2 (g) + O 2 2SO 3 (g) the standard free energies of formation of SO 3 (in K cals mole 1 ) are respectively 88.5 and 71.8. The standard free energy change, G, for the reaction (in K cals) is computed to be (a) 33.4 (b) 16.7 (c) 66.8 (d) none of these 9. Refer to Question No. 8, the thermodynamic equilibrium constant of the reaction is nearly (a) 3.0 10 16 (b) 3 10 14 (c) 3 10 6 (d) 3 10 24 10. At 298K, it was found that Ke for the reaction is 4. The G in KJ for the reaction is nearly (a) 4 (b) 2.303 8.314 298 log 4 10 3 (c) 2.303 0.821 298 log 4 10 3 (d) 2.303 1.987 596 0.301 10 3 11. Equilibrium constants K p and K c are related as (a) K p = K c [RT] n (b) K c = K p [RT] n (c) n c p ] RT [ K K = (d) K p K c = [RT] n 12. For the reaction, H 2 O(l) H 2 O(g), the vapour pressure of water at 298K is 0.6313 atm. K p for the reaction is (a) K p = pH 2 O (b) K p = 0.0313 atm (c) 3.13 10 2 atm (d) all of these 13. Refer to question no. 12, K c for the reaction can be computed using (a) 298 31 . 8 10 13 . 3 2 - (b) 2 10 13 . 3 298 31 . 8- (c) 0821 . 10 13 . 3 2- (d) 298 0821 . 10 13 . 3 2 - 2 QUEST TUTORIALS Head Office : E-16/289, Sector-8, Rohini, New Delhi, Ph. 65395439 14. The variation of Ke with temperature is given by Varit Hofbs equation of the form (a) - - = 2 1 1 2 T T T T R E ) 1 ( Ke ) 2 ( Ke log (b) - - = 2 1 1 2 T T T T R H ) 1 ( Ke ) 2 ( Ke log (c) - - = 2 1 T 1 T 1 R 303 . 2 H ) 1 ( Ke ) 2 ( Ke log (d) -...
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This note was uploaded on 10/05/2011 for the course CHEM 1201 taught by Professor Darcy during the Spring '11 term at St. Mary NE.

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Unit7 Chemical equilibrium qns - 1 QUEST TUTORIALS Head...

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