Question

at 1000 K? for the reaction 12. The equilibrium constant, K 2 HI(g) H2(g) I2(g) is 55 at 425 C. Ifa 0.40 mol sample of HI was introduced into a 1.00 mL. reaction vessel at 425 C, what are the equibrium concentrations of H2, 12, and HI? 13. Consider the following reaction which is at equilibrium at 25 C 2 NH3(g) + CO2(g) A-152.2 kJ Ni4(NH2CO2X) In which direction will the reaction shift if (a) the pressure is increased (b) the pressure is decreased (c) the temperature is increased () NH3 is removed? (e) some CO2 is added? (d) the temperature is decreased? (h) some NH4(NH2CO2) is added? (i) some Ne is added? (g) a catalyst is added ? ) the volume is increased? 14. Solid (NHa)S decomposes when heated as follows: (NHa) S() 2 NH(g) + H2S(g) a) Write expresions for Ke and Kp show the relationship between them. b) What is K, if an equilibrium mixture was found to have PNIS of 0.50 atm e) What is Ke for the reaction? 15. Write a balanced equation to show how you would make (NH4)2S. What is the name of this compound? 16. The equilibrium constant, Ke, for the reaction сосle) COg) + Cl(g) is 8.4 x 10 at 360 °c. If an emty 50.0 L vessel is charged with a 1.00 mol of COCha at 360 °C, what is the total pressure in the container reaches equilibrium?

Answer 1

(12) Write the reaction as follows: H (8) +12(8) 4 2HI(g) equilibrium constant, K. = 55 given moles of HI = 0.40 mol volume of the vessel = 1.00 mL concentration of HI. He moles of HI volume of the vessel 0.40 mol 1000 mL 1.00 mL 1L = 400 mol/L Expression for equilibrium constant is K-11 -1) ---- - - - - - - - - - - - - - - - - - - - Draw the ICE table as follows: H,(g) + 1,(g) 1(M): 0 0 C(M): +x + x E(M): +x → 2HI(g) 400 - 2x 400-2x + + Substitute the corresponding valuesin equation (1) and calculate the value of x. K K. (H) 13 (H) 55= (400 – 2x) (x)(x) 55x2 = (400-2x) 400-2x = 55x? 400-2x = 7.42x 9.42x = 400 400 x= 9.42 x= 42.5 M - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Calculate the equilibrium concentrations as follows: [H,]= +x= 42.5 M [1]= + x = 42.5M (H)= 400-2x = 400-2 (42.5 M)= 400–85= 315 M Therefore, the required solutions are is [H.]= 42.5M). [1]= 42.5 M and [H]= 315 M