Question

3. The rate of the reaction: 2CO(g) → CO2(g) + C(s) Was studied by injecting CO(g) into a reaction vessel and measuring the total pressure at constant volume.

Time (s) Ptotal (bar)

0 0.3331

398 0.3173

1002 0.2986

1801 0.2800

a) (2.5 points) Determine the partial pressure of CO at each time.

b) (4.5 point) Determine the order of the reaction and the average value for the rate constant (with the correct units).

Answer 1

a) At time = 0   we have only CO , hence pCO = 0.3331 atm  

tme = 398, CO pressure = 0.3331-2p ( see that 2CO gives 1CO2 hence 2p lost and we get 1p) , CO2 pressure = p

total P = pCO + pCO2 = 0.3331-2p+p = 0.3331-p = 0.3173 , p = 0.0158

pCO = (0.3331-2(0.0158) = 0.3015 bar

at 1002 ,   total Pressure 0.2986 = ( 0.3331-p)          ( like above )

       p = 0.0345 , pCO = ( 0.3331-2(0.0345)) = 0.2641 bar

at 1801   0.28 = 0.3331-p , p = 0.0531 bar , pCO = ( 0.3331-2(0.0531)) = 0.2269 bar

b) we can do trail and error method , we assume its 1st order and check whether it obeys 2nd order kinetics

k = ( 1/t) ( 1/p at time t 1 /p initial)

k1 = ( 1/398) ( 1/0.3015   -1 /0.3331) = 0.00079

k2 = ( 1/1002)   ( 1/0.2641 - 1/0.3331) = 0.000783

k3 = ( 1/1801) ( 1/0.2269 - 1/0.3331) = 0.00078

hence k value is nearly same , hence 2nd order kinectis obeyerd , hence order is 2

average k = ( 0.00079+0.000783+0.00078) /3 = 0.000784 bar-1 s-1