a. We know that the reverse of the reaction:left over the bang?a.H₂O(g) → H2(g) + 1/2O2(g)a finite equilibrium constant of 8.581×10-4¹ at 298 K.How could there be any hydrogen or oxygen gasAmount at equilibriumMole fractionsfractions belowconstant 1 bar, as such we don't need to worry about partial pressures etc.0₂fill out the remaining entries of the table of moleLet's assume the pressure is maintained at aH₂On(1-a)H₂na4x³(2 + a)(2-2a)²b. Using the data above please write the expression for Kx- using a.c. Next, set this expression equal to the known equilibrium constant 8.581×104¹ and solve for a.I suggest you use the on-line tool Wolfram solver. For instance, if you were trying to solve thisexpression:(XHz)(Xo,)/2XH₂O= 7.36 x 10-81you would type in the following: 4*x*x*x/(2+x)/(2-2*x)/(2-2*x) =7.36e-81. Solving it wouldgive a=2.451×10-27.d. Given the value of a you got in pt. c, now calculate the amount of H₂ and O₂ you have using~na. However, do not use n=1 mol, rather, use Avogadro's number n=6.02×10²3 molecules. Thisrepresents the number of hydrogen and oxygen molecules in equilibrium with a mole of water.Now can you explain that the reaction is indeed very strongly slanted to the reactant (water)side?

The equilibrium expression is given as, a) H2O ↔ H2(g)+12O2(g)at t=0 n 0 0at t=tequi n(1-α) nα nα/2Total no of moles = (n-nα+nα+nα2)=n+nα2Mole fraction of H2O = No of moles of H2OTotal moles=n-nαn+nα2= 1-α1+α2Mole fraction of O2 = No of moles of O2Total moles=nα/2n+nα2=α21+α2Mole fraction of H2 = No of moles of H2Total moles=nαn+nα2= α1+α2 b)Kχ=χH2χO212χH2O =α(1+α2)α2(1+α2)121-α1+α2 C) C) Now, Kχ=χH2χO212χH2O =α(1+α2)α2(1+α2)121-α1+α2=8.581×10-41Or, 4α3(2+α)2-2α2=7.36×10-81 4α3(2+α)4(1-2α+2)=7.36×10-81α3=(2+α)×(1-2α+2)×7.36×10-81α3 = -2α2-α+6 (7.36×10-81) =-1.47×10-80α2-7.36×10-81α+44.16×10-81α=2.451×10-27