4. Substance A decomposes according to the reaction: 0.200 mole of the substance

Question

4. Substance A decomposes according to the reaction: 0.200 mole of the substance A is added to a 400-ml evacuated container at 50 C. Some of the material was removed every 2.50 minutes listed in the table below Time Concentration of A (minutes) (moles/liter) 2.50 5.00 7.50 0.5000 M 0.4375 M 0.3750 M 0.3125 M 12.5 0.1875 M 15.0 0.1250 M Various graphs were prepared from the data by plotting [A] vs time, In [A] vs time, and 1MA] vs time. 2.0 0.8 Time (minutes) 05 (a) Using the drawings above as a guide, what is the order of the reaction? Oorder (b) Determine the half-life of the reaction. 10min calculate its rate constant,complete with (d) Calculate the initial rate thereaction tra (c) Give the rate law for the reaction and the correct units..025 Wmh, rote .025 -0) How does the valueati:0)compare with the initial rate whe t-15) same What would be the concentration of substance B after 20 minutes? Time (minutes) 0 c) , 437 -k(a5)t.

Explanation / Answer

d) For a zero order reaction,

-dCA/dt = K,

On integrating

CA = -Kt + c

at t = 0; CA = CA0;

CA0 = -K*0 + c

c = CA0,

On substituting and rearranging,

CA0 - CA = Kt

For instantaneous rate, we draw a tangent from the point t=0 in a CA Vs time plot, extending towards y axis. The slope of this line gives the instantaneous rate. As the plot of our problem is already a straight line, the slope is given as,

Instantaneous rate = (0.5 - 0.125) / (15 - 0 ) = 0.025 mol/litre.sec

The instantaneous rate at t = 15 mins is same as the that at t = 0 min. From our rate equation it is evident that the reaction rate is not a function of concentration. [ ra = -dCA/dt = k ]

e) CA0 - CA = Kt

CA0 = 0.5 mol/litre

k = 0.025 mol/litre.sec

t = 20 min

CA = 0.5 - 0.025*20 = 0 moles/litre . This means the reacton is complete.

For cases where, t >= CA0/k, CA = 0. In our case CA0/k = 0.5/0.025 = 20 min, thus CA = 0.

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