I am having trouble figuring out questions 7-10b please help. If you answered -y
ID: 498954 • Letter: I
Question
I am having trouble figuring out questions 7-10b please help. If you answered -yes" to say part of the previous question, what would be the slope of the line? Consider the function k = Ae^(-E/RT) Would a graph of k versus E be a straight line? If not, how could how make a linear graph that shows the relationship between k and E? In chemistry lecture next week, you will study integrated rate laws. These are equations that relate the concentration of a reactant to time. As shown in the table below, the form of the integrated rate law depends on the reaction order. For this weeks lab, you do not need lo know the details of these equations, but you do need to apply the skills that were used in the previous questions, and the information in the table, to determine bow to generate a linear graph based on these integrated rate laws. Rate laws and integrated rate laws for 0, 1^st, and 2^nd order reactions. In the integrated rate laws, [A]_0 refers to the initial concentration of A (at time t = 0). Imagine you measure the concentration of a reactant [A] at several limes during a reaction. a If you plot [A] versus t and see a straight line, what is the reaction order with respect to A? b. What would be the slope of this line? Imagine you measure the concentration of a reactant [A] at several times during a reaction a. If you plot 1/[A] versus t and see a straight line, what is the reaction order with respect to A? b. What would be the slope of this line? Imagine you measure the concentration of a reactant [A] at several times during a reaction a. If the reaction were first order with respect to A, what variables would you need to plot in order to get a linear graph? b. What would be the slope of this line?Explanation / Answer
Answer
8) from the given integrated rate law y = mx + c format in which y = [A] and x = t and m = - k
a) zero order
b) -k
9) from the given integrated rate law y = mx + c format in which y = 1 / [A] and x = t and m = k
a) second order
b) k
10) if it is first order integrated rate law is
ln ([A]) = - kt + ln ([Ao])
y = m x + c
so graph should be drawn
a) ln ([A]) vs t
b) slope will be - k
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