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Chapter 58: Problem 3

If one obtained data for a second-order reaction, and then made a graph of \(1 /(\mathrm{R})\) versus \(t\), the resulting plot would be a straight line. a) What would be the slope? b) What would be the \(y\) -intercept?

Short Answer

Expert verified
a) The slope of the graph would be the rate constant for the reaction. b) The y-intercept of the graph would be the reciprocal of the initial rate of the reaction.

Step by step solution

01

Understanding the Second-Order Reaction

A second-order reaction is a form of chemical reaction that proceeds at a rate proportional to the square of the concentration of a single reactant or to the product of the concentrations of two reactants. The general form of the rate law for second-order reactions is \(r=k[A]^2\), where \(r\) is the rate of reaction, \(A\) is the concentration of a reactant and \(k\) is the rate constant.
02

Slope of the Graph

The slope of the graph represents the rate constant \(k\) for the reaction. This is because in the case of a second-order reaction, the relationship between \(1/R\) and \(t\) is linear and the slope of this line equals to the rate constant \(k\).
03

Y-Intercept of the Graph

The \(y\)-intercept of the graph represents the initial rate of the reaction, or specifically, the reciprocal of it. This is because when \(t=0\), \(1/R\) is equal to \(1/R_0\) where \(R_0\) is the initial rate of the reaction.

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Most popular questions from this chapter

If one obtained data for a first-order reaction, and then made a graph of \(\ln (\mathrm{R})\) (along the vertical axis) versus \(t\) (along the horizontal axis), the resulting plot would be a straight line. a) What is the slope? b) What is the \(y\) -intercept? (Hint: Compare the first-order integrated rate law to the equation \(y=m x+b\).)

Recall that for a first-order reaction: $$ \ln (\mathrm{R})=\ln (\mathrm{R})_{0}-k t $$ a) When \(t=t_{1 / 2}\), what is the value of \((\mathrm{R})\) in terms of \((\mathrm{R})_{0}\) ? b) Show that \(t_{1 / 2}=\frac{\ln 2}{k}=\frac{0.693}{k}\) for a first-order reaction.

Recall that for a second-order reaction: $$ \frac{1}{(\mathrm{R})}=\frac{1}{(\mathrm{R})_{\mathrm{o}}}+k t $$ a) When \(t=t_{1 / 2}\), what is the value of \((\mathrm{R})\) in terms of \((\mathrm{R})_{0}\) ? b) Show that \(t_{1 / 2}=\frac{1}{k(R)_{0}}\) for a second-order reaction.

For both integrated rate laws, show that (R) decreases as \(t\) increases.
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