At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3 At this temp, 0.3000 mol of H2 and 0.300 mol o f I2 were placed in a 1L container to react. What concentration of HI is present at equilibrium? H2+I2<-> 2HI
1 answer:
Your answer is: Kc = [HI]² / {[H2] [I2]} <span>53.3 = (2x)² / {(0.400M - x)(0.400M - x)} </span> <span>sqrt(53.3) = 2x / (0.400M - x) </span> <span>2.92 - 7.30x = 2x </span> <span>x = 0.314M </span> <span>[HI] = 2x = 0.628M :) hopefully this helped. (:</span>
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