Question

any help is appreciated

Question 16 (5 points) If the Keq of the following reaction is 0.38, 2A (9) + 3B (s) = 7C (), what is the Keq of the reaction below? 14C (0) = 4A (9) + 6B (s) 5.26 O 0.760 1.12 6.93 0.144 Question 17 (5 points) Al 105°C. AsEs decomposes into ASF3 and F2 according to the following chemical equation.

Answer 1

SOLUTION 160 The given chemical equation is 2A (g) + 3B (s) 766 Keg for the above reation is 0.28. Kea NOTE : When the sequilibrium reaction is reversed the new keg that is, keg becomes 1 , where keg represent equilibrium constant of initial reacton. for the reaction, 70 (1) = A (g) + 3B (s) k'eqo Konobe * 2.63 Also, when the chemical reaction is multiplied by a factor the new equilibrium Constant will be increased by that power . for reaction, 140 (1) 4A(g) + GB (s) We notice that the coefficient have been multiplied by 2 from second case. So the new equilibrium constant, k'eg = (keq) = (163) ² = 6.93

reaction is : reaction is given equilibriun In the given chemical reaction ! Asfy (9) Ast (g) + f (9) We know that the relation between and kp sor a given equilibrium kp = kc (Rijan | where kpake are equilibrium constants and R is the universal gas constant. I represents temperature of system (in kelvin) An represents the difference in number of gaseous element moles from product to reactout side biven - Ke = 3.3x10-3 M R = 0.082 adm litre moltkt T- 20 105°c or (105+273) K = 378k. An= (1+1) -1 -1 100 product side of has 2 gase molest and reactant side mole 2. Kp= 3.3x10 3 +(0.082 * 378)' 0.10 or ke - 1.0 x 10-1

Hope that this will help you.

Kindly upvote it.