Solve this:

Solve this: 20. 21 22. t24. 25. 29. 30 CHEMICAL EQUILIBRIUM For the reaction pct. (g) + , the value of Kc at 2500 c is 26. Kp at this temperature Will be D) 0.46 C) 0.83 A) 0.605 , B) 0.57 One mole of nitrogen was mixed with 3 moles Of hydrog:n in a closed 3 litre Vessel. nitrogen is converted into NH3. Then Kc for the — N2 is A) 0.36 litre mole-I B) 0.46 litre mole-I C) 0.5 mole-I litre D) 0.2 mole-I litre For the following reaction at 2500C, the value Of is 26, then the value Of Kp at the temperature Will be 4 Ciao C) 0.83 B) 0.605 A) 0.57 D) 0.91 At constant temperature 80% NO dissociates into Nz and 02 the equilibrium constant for 2NOe N, +0, is C) 16 D) 0.25 If x is the fraction of NH, dissociated at equilibrium in the reaction 2NHa(g) eN2(g) +3H2(g): then starting with 2 moles of NH, the total 'number of moles reactants and products at equilibrium is B) 2+2x D) 4 + 2x The reaction proceeds to right hand side to the extent of 99.9%. What is the equilibrium constant for the reaction? C(g) + D (g) A (g) + B(g) is approximately A) 106 C) 104 If one third of HI decomposes at a particular temperature, KC for + La is A) 1/16 B) 1/4 C) 1/6 25g of N2 and 6g of H2wcre mixed. At equilibrium 17g of NH3 was formed. The weight of N2 and H2 of equilibrium are respectively C) lag, 3g A) 1 lg,zero The equilibrium constant for the reaction 2X(g) + Y is 2.25 litre mole¯l . What would be the concentration of Y at equilibrium with 2.0 moles of X and 3.0 moles of Z in one litre vessel ? C) 2.0 M B) 2.25 M A) 1.0 M D) 4.0 M 5. 6. 7. 8. 9. One mole of nitrogen was mixed with 3 moles of hydrogen in a closed 3 litre vessel. 20% Of nitrogen is converted into NIL. Then Kc for the Y2N2 + % H2 is ) 0.36 litre mort B) 0.46 litre mort C) 0.5 mole' litre D) ().2 mol"t litre 1.1 mole of A is mixed with 2.2 mole of B and the mixture is then kept in one litre flask fill 2C + D. At the equilibrium 0.2 mole of C are formed. the equilibrium is attained A + 2B The equilibrium constant of the reaction is C) 0.003 B) 0.002 A) 0.001 D) 0.004 18. the

Dear Student,

                      12N2          +           32H2                              NH3Initial              1                               3                                           0At equil.      1-20100                   3-3×202×100                         20×2100Kc = [NH3][N2]12[H2]32     =20×2100×3(80100×3)12(240100×3)32     =0.36Hence, correct answer is (A)Kindly ask the remaining query in the next thread. 

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