Question

For which of the following reactions is the product formation favored

9. Consider the following reactions. In which cases is the equilibrium unaffected by pressure? ΔΗ0-206.2 kJ AHo 172.5 k] He =-9.4 kJ ΔΙΗ0-541 kJ лно-66.4 kJ 1) CO(g) + 3H2(g) 2) CO2(g) + C(s) F 3) H2(g)+ L2(g) 2HI(g) CH4(g) + H2O(g) 2CO(g) 5) N2(g)+202(g)2NO.(g) a. 1,5 b. 2,5 c. 3, 4 d. 1, 2, 5 e. 2, 3,4 10. For which of the following reactions is product formation favored (i.e, drive equilibrium to the right) by pressure and low temperature? a. No(g)+3H.(g)2NH(g) b. 2CO (g) 2CO(g)+O2(8) c. 30() 20 (8) d. Br()+ 2NO() 2NOBr(g) e. N(g)+ 20(g)2NO(g) /Ho =-91.8 kJ AHo- 566 kJ ΛΙ 10-285 kJ AHo 66.4 k

Answer 1

(3), (4) equilibrium won't be affected by pressure. (3) H_2(g) + I_2(g) rightleftarrow 2 HI(g), Delta H^0 = -9.4 kJ (4) H_2(g) + F_2(g) rightleftarrow 2 HF(g), Delta H^0 = -541 kJ (As per Le - Chatelier's principle) By changing pressure, The above two equilibrium won't be affected as Delta_ng = 0 (Delta_ng = sigma gaseous moles of products - sigma gaseous moles of reactants) part (c) is correct. The following equilibrium product favoured at high pressure & low temperature is: - (a) N_2(g) + 3H_2(g) rightleftarrow 2NH_3(g), Delta H^0 = -91.8 kJ since the above reaction is exothermic (heat is released as Delta H^0 is negative) so on lowering the temperature reaction will proceed in that direction where temperature is high i.e forward direction by Le-chatelier's principle. \r\n Also an increasing the pressure, volume decreases so, the reaction will proceed in that direction in which no. of gaseous moles are less as per Le-Chatelier's principle i.e product side.
the above image talks about temperature and pressure effects on

Equilibrium . ( Le - Chatelier's principle )