Homework Answers
A. This is a second order reaction. The reason is in 2nd order reactions the 1/A vs t graph will be straight line.
B. The value of A0 is 6.5789. According to the equation of 2nd order reaction,
At = -k t + A0
3. The rate constant is 5.741. This is the slope of the straight line in above graph. The unit of the rate constant of a 2nd order reaction as like in this case is Mol-1. L. s-1
4. The half life of the second order reaction is given below,
T1/2 = 1/k. A0
Have to substitute the values of k and A0 in above equation.
= 1/ 5.741×6.5789
= 0.026 min.
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I need help with A & B.
The graph for both are on the right.
I need help with part C.
I need help with part D as
well.
So all together Parts A, B, C, D.
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A substance (A) reacts to form another substance (B): 3A(g) --> 2B(g) The reaction is run at a particular temperature with the concentrations of A and B monitored over time and plotted in the graph. At what time was equilibrium first reached and what is the approximate value of the equilibrium constant? [B] concentration/M (A) 0 10 20 30 40 50 60 70 80 90 time / min 100 110 120 130 140 150 0 Equilibrium was reached at 70...
I need help with A & B.
The graph for both are on the right.
I need help with part C.
I need help with part D as
well.
So all together Parts A, B, C, D.
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