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# At 700 K, CO2 and H2 react to form CO and H2O. For this process Kc is 0.11. If a mixture of 0.45 mole of CO2 and 0.45 mole of H2 is heated at 700K.(i) Find out the amount of each gas at equilibrium state.(ii) After equilibrium is reached another 0.34 mole of CO2 and 0.34 mole of H2 are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state.

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Solution

## CO2(g)+H2(g)⇌CO(g)+H2O(g)Initial no. of moles: 0.45 0.45 0 0No. of moles at eqm: (0.45−x) (0.45−x) x xApplying law of mass action,Kc=[CO][H2O][CO2][H2]=x×x(0.45−x)(0.45−x)=0.11So, x(0.45−x)=0.33x=0.11At equilibrium, [CO2]=[H2]=0.34 mole[CO]=[H2O]=0.11 mole(ii) CO2(g)+H2(g)⇌CO(g)+H2O(g)Initial moles 0.34 0.34 0.11 0.11 +0.34 +0.34 − − =0.68 =0.68 0.11 0.11At equilibrium:(0.68−y), (0.68−y), (0.11+y), (0.11+y)Kc=[CO][H2O][CO2][H2]=(0.11+y)(0.11+y)(0.68−y)(0.68−y)=0.11or y=0.086At equilibrium, [CO2]=[H2]=0.11+0.086=0.196 mole[CO]=[H2O]=0.68−0.086=0.594 mole

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