Question

3. At a particular temperature, K. = 378 for the reaction shown below. In an experien 2.45 mol NO are introduced into a 2.00-L flask. a) Set up the complete ICE table for solving this equilibrium problem. = N2(e) + O2(g) 2NO() 1,275 2 1.225 b) Show the equilibrium constant expression with the expressions involving x from the ICE table substituted. Note that I am not asking you to solve the equilibrium problem. You're just asked to set the problem up as far as showing the equilibrium constant expression with x included as if you were about to solve it. (products

Answer 1

Equilibrium concentrations also provided for you. Thank you.
& Grivenik. - 378 Initial concentration (No = 2.45 und of - - 1.0225M 2 L ICE ~ Initial concentration, Eduilibrium TICE table] 2NO(g) N (g) + O (9) H kpeq 1.225 + x X U - E - 2x 1.225-2x + x x x Eduilibrim Constanta (products at edo reactantia in Ke= (^2} {0} [no] kea (1.225~2x)2
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solving & Fraud : & ka 378= 72 (1.225_2x)2 => 39664x?– 4,98 +1.500625)= ** = 1511x² - 1852.28 +567.23625 solve quatratic equation: що, : х. —Ь+ — 4-с La gives: x=0.597143 \ in at suitabrium No= 1.225-2x=0.038714 (^2= x = 0.5971 43 23 = x= -897143