Time (s) [NOBr]( M ) 0.0e+00 0.100 5.0e+01 0.0172 1.0e+02 0.00939 1.5e+02 0.0064

Question

Time (s) [NOBr](M) 0.0e+00 0.100 5.0e+01 0.0172 1.0e+02 0.00939 1.5e+02 0.00646 2.0e+02 0.00493 1 a) This reaction issecond-order with the rate constant k = 5.79e+00M-1 s-1.
b) This reaction issecond-order with the rate constant k = 9.65e-01M-1 s-1.
c) This reaction issecond-order with the rate constant k = 3.22e-01M-1 s-1.
d) This reaction isfirst-order with the rate constant k = 4.83e-01s-1.
e) This reaction issecond-order with the rate constant k = 3.86e+00M-1s-1.
1 a) This reaction issecond-order with the rate constant k = 5.79e+00M-1 s-1.
b) This reaction issecond-order with the rate constant k = 9.65e-01M-1 s-1.
c) This reaction issecond-order with the rate constant k = 3.22e-01M-1 s-1.
d) This reaction isfirst-order with the rate constant k = 4.83e-01s-1.
e) This reaction issecond-order with the rate constant k = 3.86e+00M-1s-1.
1 Time (s) [NOBr](M) 0.0e+00 0.100 5.0e+01 0.0172 1.0e+02 0.00939 1.5e+02 0.00646 2.0e+02 0.00493

Explanation / Answer

If you plot ln [NOBr] vs. t and you get a straight line, then it isfirst order If you plot 1/[NOBr] vs. t and you get a straight line, then it issecond order In this case it is a second order rxn the slope of the line is the rate constant k slope = 0.9647 rate constant = 0.9647 Answer is B

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