To understand how to use integrated rate laws to solve for concentration. A car
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To understand how to use integrated rate laws to solve for concentration. A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours? This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145. 55 mi/hr times 2 hr = 110 miles traveled milemarker 145 - 110 miles = milemarker 35 The rate constant for a certain reaction is k = 4.80 times 10^-3 s^-1. If the initial reactant concentration was 0.950 M, what will the concentration be after 20.0 minutes? A zero-order reaction has a constant rate of 3.10 times 10^-4 M/s. If after 40.0 seconds the concentration has dropped to 6.50 times 10^-2 M, what was the initial concentration?Explanation / Answer
The units of rate constant suggest the reaction to be 1st order. For 1st order reaction,
-dCA/dt= KCA, CA= concentration of reactant and K is rate constant.
When the equation is integrated noting that CA= CAO( initial concentration ) and CA= CA at any time t.
CA= CAO*e(-Kt)
Given K= 4.8*10-3/sec and t=20minutes= 20*60 sec=1200 sec and CAO=0.95M
CA= 0.95*e(-4.8*1200/1000)= 0.002994M
2. for zero order reaction, -dCA/dt= K
When the equation is integrated noting that CA= CAO( initial concentration ) and CA= CA at any time t. K is rate constant
CA= CAO-Kt, CAO=CA+Kt, CAO= 6.5/100+40*3.1/10000 = 0.0774N
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