Integrated rate law second order

The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b y m x+b . The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b
.
The reactant concentration in a second-order reaction was 0.630
M after 270 s and 3.20×10−2M after 870
s . What is the rate constant for this reaction?
Express your answer with the appropriate units. Indicate the
multiplication of units, as necessary, explicitly either with a
multiplication dot or a dash.
Please explain how to complete this probelm. We havent gone over
this in class, yet we have this assignment and I am so lost! 
t Using Integrated Rate Laws The integrated rate laws for zero-, first-, and second order reaction may be arranged such that they resemble the equation for a straight line, mac b Order Integrated Rate Law Graph Slope vs. k The reactant concentration in a second-order reaction was 0.630
M after 270 s and 3.20×10−2M after 870
s . What is the rate constant for this reaction? M −2 M Express your answer with the appropriate units. Indicate the
multiplication of units, as necessary, explicitly either with a
multiplication dot or a dash. Please explain how to complete this probelm. We havent gone over
this in class, yet we have this assignment and I am so lost! t Using Integrated Rate Laws The integrated rate laws for zero-, first-, and second order reaction may be arranged such that they resemble the equation for a straight line, mac b Order Integrated Rate Law Graph Slope vs. k

Since reaction is second order. We know from second order reaction 1/[A] = Kt + 1/ [A]_0 equation (1) where [A]_0 = Initial concentration of reactant A [A] = Concentration of A after time ‘t’ K = Rate Constant Given t_1 = 270s, [A] = 0.630 M t_2 = 870 s, [A] = 3.20 times 10^-2 M = 0.032 M We can write (1) in terms of t_1 and t_2 1/[A]_t_1 = kt_1 + 1/[A]_0 equation (2), 1/[A]_t_2 = Kt_2 + 1/[A]_0 equation (3) (2) – (3), 1/[A]_t_1 – 1/[A]_t_2 = Kt_1 – K_t_2 = K(t_1 – t_2) = K (270S – 870S) 1/0.630 M – 1/0.032M = K(-600S) implies 1.587M^-1 – 31.25 M^-1 = -K times 600S K = 29.66 M^-1/600S = 0.0494 M^-1 S^-1 = 4.94 times 10^-2 M^-1 S^-1 Rate constant = 4.94 times 10^-2 M^-1 S Ans
The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b
.
The reactant concentration in a second-order reaction was 0.630
M after 270 s and 3.20×10−2M after 870
s . What is the rate constant for this reaction?
Express your answer with the appropriate units. Indicate the
multiplication of units, as necessary, explicitly either with a
multiplication dot or a dash.
Please explain how to complete this probelm. We havent gone over
this in class, yet we have this assignment and I am so lost! 
t Using Integrated Rate Laws The integrated rate laws for zero-, first-, and second order reaction may be arranged such that they resemble the equation for a straight line, mac b Order Integrated Rate Law Graph Slope vs. k

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