Question

please solve

The equilibrium constant, K, for the following reaction is 4.85x10-2 at 641 K. COC12(8) CO(g) + Cl2(g) An equilibrium mixture of the three gases in a 6.59 L container at 641 K contains 0.205 M COCI, 9.98x102 M CO and 9.98*10-M CI2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 15.5 L? [COCI2] = [CO] [Cl] Submit Answer

Answer 1

Solution Given Data K: 4-85X10" T 641K Volume of Container = 6.59 L I coug] = 0. 205M Croj - 9.98X10*? M TC12] : 9.98x102M - Now, the contente volume of container increases opto 15.5 L: - Collacg, - Corg) + Cl cg). concentration = - No. of moles of solute Volume of solution (L.) le . M= n VOL) na Mx VCL) ncoce = 0.205 MX 6.591 = 1.35 mol nco = 9.98x102M X 659 L = 0.658 moll nag= 9.98810² M x 6.592= 0.658 mol Scanned with CamScanner

Now, these 15.52 hese moles Container. So more . of the gases dissolved in concentration 7 M= Concentration A No of moles of solute volume of solution (2) [cou] = 1.35 mol 15.5L -0.087 M . [col= 0.65 mol 15-5.L = 0.042 M. tul, J = 0.658 mo) 15:52 = 0.042 M Scanned with CamScanner

Cocoa 10.087 IM ICO]= 0.042 IM [4] = 0.042 ]M cs Scanned with CamScanner