Question

Suppose a 500. mL flask is filled with 1.5 mol of CO, 1.8 mol of H,O and 0.60 mol of CO,. The following reaction becomes possible: CO(g) +H2O(g) + CO2(g)+H2(g) The equilibrium constant K for this reaction is 3.75 at the temperature of the flask. Calculate the equilibrium molarity of 1,0. Round your answer to two decimal places. IM | xs ?

Answer 1

Equilibrium molarity of H2O is

1.64 M
since, Ang ire difference in noi of moles of gaseoul products & reactants is zerosol kn=kcZkp ence, it doeen't tepents upon volume cocgl + H₂0 g) 1.5 8 1. -n -n (1.5-1) (1.8-x 0 Congo + Hig) .60 - (Initially ta t a (Change (6 60+x) tu (At ) soy 2 K=hco, H₂ ... hco. nuo 3.75 = (0.60+2) n. (1.5-n (1.8.7%) 3.75 = o.font x² 2.7 -3.3n+ x² 10.125 - 12.375n+ 3.75 n² = o.font ut (2751² - 12.925 n +10.125=0 = = So By quadratic formula x= +12.975 I C-12.975)2 - 4x2.758 (0.125 2x2.75 = 12.975 + √56.975 = 12.975 t 7,548 5.5 sis

x = 12.975 t 7,548 (OR) 2= 72.975-7.548 = 3.73 COR) x = 0.98 As, x=3.73 negative soy mate the value of mole & conc? correct value is = 0.98 Hence, Equilibrium molarity of Ho is 118-2 = 108-0-98 soox 103 [H] 1.64M /