1.75 moles of H2O2 WERE PLACED IN A 2.50-L REACTION CHAMBER AT 307 DEGREES CELSI

Question

1.75 moles of H2O2 WERE PLACED IN A 2.50-L REACTION CHAMBER AT 307 DEGREES CELSIUS. AFTER EQUILIBRIUM WAS REACHED, 1.20 MOLES of H202 REMAINED. CALCULATE THE EQUILIBRIUM CONSTANT, Kc, for the reaction 2H202---->2H2O+O2
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Explanation / Answer

2mol H2O2(g) ---> 2 mol H2O(g) & 1 mol O2(g) 1.75 mol ..........initially 1.20 moles ......at equilibrium remain, so we lost 0.55 moles of H2O2 in a shft to the right--> 0.55 mole ---> --> 0.55 mol H2O & half as many O2: 0.275 mol O2 at equilib find molarities H2O2 = (1.20 mol) / (2.50 L) = 0.48 Molar H2O = (0.55 mol) / (2.50 L) = 0.22 Molar O2 = (0.275 mol) / (2.50 L) = 0.11 Molar Kc = [H2O]^2 [O2] / [H2O2]^2 Kc = [0.22]^2 [0.11] / [0.48]^2 Kc = (0.0484] [0.11] / (0.2304) Kc = 2.3 X 10^-2

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