Nitric oxide and bromine at initial pressures of 98.4 and 41.3torr, respectively

Question

Nitric oxide and bromine at initial pressures of 98.4 and 41.3torr, respectively, were allowed to react at 300. K. At equilibriumthe total pressure was 110.5 torr. The reaction is as follows. 2 NO(g) + Br2(g) 2NOBr(g) (a) Calculate the value ofKp. 1 (b) What would be the partial pressures of all the species if NOand Br2, both at an initial pressure of 0.30 atm, wereallowed to come to equilibrium at this temperature? PNO 2 atm PBr2 3 atm PNOBr 4 atm I got as far as finding values for ICE 2NO BR2 2NOBr I 98.4 41.3 0 C -x -2x +2x E 98.4-x 41.3-2x 2x kp= (2X)^2 / (98.4-2x)(41.3-x)

Explanation / Answer

a) 23.20587251 for b the ice table should look like this .30        .30          0 -2x        -x          2x .30-2x    .30-x      2x 2x/(.30-2x)*(.30-x)=ksp from problem a solve for x and then find the vaules at eqb by pluging it back intothe equation

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