Chapter 13, Problem 13.142QP

(a)

Interpretation Introduction

Interpretation:

The rate constant for the growth in the number of transistor on an integrated circuit has to be determined using the given plot ln N versus year.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

The rate law expresses the relationship of the rate of a reaction to the rate constant.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

For first order reaction, $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.

Moore’s law states that the number of transistors per square inch on integrated circuits had doubled every year since their invention (1958).

Explanation of Solution

Given plot of ln N versus t (year) is shown below,

Figure 1

The plot of ln N versus t is linear for a process which follows first order kinetics. And so the given process follows first order kinetics.

The rate can be described using the equation,

$Rate=\frac{\Delta N_t}{\Delta t}=k{N}_t$

Where N is the number of transistor on an integrated circuit, which is roughly doubles every 1.5 year according to the Moore’s law.

For first order reaction, $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.

For this case, the equation can be rearranged as follows,

$\ln N_t=kt+\ln N_0$

Comparing this equation to the straight line equation ($y=mx+c$) thus, the slope of the line will give rate constant and it is $0.40year{}^{-1}$

$\begin{array}{c}\text{slope}\left(\text{m}\right)\text{=}\frac{\text{Δy}}{\text{Δx}}\text{=}\frac{\text{14-10}}{\text{1990-1980}}\\ \text{=}\frac{\text{4}}{\text{10}}\text{=}\text{0}\text{.40}{\text{year}}^{\text{-1}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The time required for $N_t$ to double has to be determined using the rate constant value.

Concept introduction:

The rate law expresses the relationship of the rate of a reaction to the rate constant.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

For first order reaction, $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.

Moore’s law states that the number of transistors per square inch on integrated circuits had doubled every year since their invention (1958).

Explanation of Solution

Given plot of ln N versus t (year) is shown below,

Figure 1

The time required for $N_t$ to double can be determined as follows,

For first order reaction, $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.

$k=0.40year{}^{-1}$

$\begin{array}{c}\text{ln}\frac{{\text{2N}}_{\text{t}}}{{\text{N}}_{\text{t}}}\text{=}\left(\text{0}\text{.40}\text{year}{}^{\text{-1}}\right)\text{t}\\ \text{ln}\text{2}\text{=}\left(\text{0}\text{.40}\text{year}{}^{\text{-1}}\right)\text{×}\text{t}\\ \text{0}\text{.693}\text{=}\left(\text{0}\text{.40}\text{year}{}^{\text{-1}}\right)\text{×}\text{t}\\ \\ t=1.7years\end{array}$

This value is very close to the value mentioned in Moore’s law.

(c)

Interpretation Introduction

Interpretation:

The number of transistors on an integrated circuit  $N_t$ in the year of $2100$ has to be determined with respect to the Moore’s law.

Concept introduction:

The rate law expresses the relationship of the rate of a reaction to the rate constant.

Rate equation for the general reaction $\text{A+B}\to \text{Product}$ is,

$Rate=\underset{\begin{array}{l}rate\\ constat\end{array}}{k}\left[A\right]\left[B\right]$

Order of a reaction:  The sum of exponents of the concentrations in the rate law for the reaction is said to be order of a reaction.

For first order reaction, $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

$\left[A\right]$ is the concentration of reactant A at time t ${\left[A\right]}_0$ is the  initial concentration of reactant, k is the rate constant.

Moore’s law states that the number of transistors per square inch on integrated circuits had doubled every year since their invention (1958).

Explanation of Solution

Given plot of ln N versus t (year) is shown below,

Figure 1

The time required for $N_t$ to double can be determined as follows,

For first order reaction, $-kt=\ln \left(\frac{\left[A\right]}{\left[A_0\right]}\right)$

For this case, the equation can be rearranged as follows,

$\ln N_t=kt+\ln N_0$

Assume the year 1960 as $t=0$ and taking $\ln N_0=2$

The year 2100 would corresponds to $\text{t}=\text{2100}\text{-1960}=\text{140}\text{years}$

Substituting known values in the above mentioned equation,

$\begin{array}{c}\text{ln}{\text{N}}_{\text{t}}\text{=}\left(\text{0}\text{.40}\text{year}{}^{\text{-1}}\right)\left(\text{140}\text{year}\right)\text{+}\text{ln}\text{2}\\ \text{ln}{\text{N}}_{\text{t}}\text{=}\text{56}\text{.69}\\ {\text{N}}_{\text{t}}\text{=}{\text{e}}^{\left(\text{56}\text{.69}\right)}\\ {\text{N}}_{\text{t}}\text{=}{\text{4×10}}^{\text{24}}\end{array}$

Thus, there will be $4\times {10}^{24}$ transistors on a circuit in the year 2100. This number is unrealistic. In reality, many scientists believes that we are already approaching the end of Moore’s law due to quantum mechanical limits to the number of transistors that can be placed on a single circuit.