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Chapter 11: Problem 66

The chirping rate of a cricket \(\mathrm{X}\), in chirps per minute near room temperature, is $$\mathrm{X}=7.2 t-32$$ where \(t\) is the temperature in \({ }^{\circ} \mathrm{C}\). (a) Calculate the chirping rates at \(25^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). (b) Use your answers in (a) to estimate the activation energy for the chirping. (c) What is the percentage increase for a \(10^{\circ} \mathrm{C}\) rise in temperature?

Short Answer

Expert verified
Question: The chirping rate of a cricket is given by the formula X = 7.2 * t - 32, where X is the chirping rate, and t is the temperature in Celsius. Find the percentage increase in the chirping rate for a 10°C rise in temperature, if the initial temperature is 25°C. Answer: To solve this problem, first, calculate the chirping rates at 25°C and 35°C using the given formula. Next, find the difference between the chirping rates at these temperatures. Finally, calculate the percentage increase in chirping rate for a 10°C rise in temperature using the formulas mentioned above.

Step by step solution

01

Calculate the chirping rate at \(25^{\circ} \mathrm{C}\)

To find the cricket's chirping rate at \(25^{\circ} \mathrm{C}\), we need to substitute t with 25 in the given formula X = 7.2 * t - 32: $$\mathrm{X} = 7.2\times25 - 32$$
02

Calculate the chirping rate at \(35^{\circ} \mathrm{C}\)

To find the cricket's chirping rate at \(35^{\circ} \mathrm{C}\), we need to substitute t with 35 in the given formula X = 7.2 * t - 32: $$\mathrm{X} = 7.2\times35 - 32$$
03

Find the difference in chirping rates

The activation energy can be estimated by finding the difference between the chirping rates at the two temperatures: $$\mathrm{Difference} = \mathrm{Chirping\,Rate}\,_{35^{\circ} \mathrm{C}} - \mathrm{Chirping\,Rate}\,_{25^{\circ} \mathrm{C}}$$
04

Calculate the percentage increase for a \(10^{\circ} \mathrm{C}\) rise in temperature

To find the percentage increase in chirping rate for a \(10^{\circ} \mathrm{C}\) rise in temperature, we first need to find the relative increase in chirping rate and then use it to calculate the percentage increase: $$\mathrm{Relative\,Increase} = \dfrac{\mathrm{Difference}}{\mathrm{Chirping\,Rate}\,_{25^{\circ} \mathrm{C}}}$$ $$\mathrm{Percentage\,Increase} = \mathrm{Relative\,Increase} \times 100$$

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Most popular questions from this chapter

At low temperatures, the rate law for the reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)$$ is as follows: rate \(=\) constant \(\times\left[\mathrm{NO}_{2}\right]^{2}\). Which of the following mechanisms is consistent with the rate law? (a) \(\mathrm{CO}+\mathrm{NO}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}\) (b) \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4} \quad\) (fast) \(\mathrm{N}_{2} \mathrm{O}_{4}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{NO} \quad\) (slow) (c) \(2 \mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\) NO \(\quad\) (slow) \(\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} \quad\) (fast) (d) \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2} \quad\) (slow) \(\mathrm{O}_{2}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2} \quad\) (fast)

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The greatest increase in the reaction rate for the reaction between \(\mathrm{A}\) and \(\mathrm{C}\), where rate \(=k[\mathrm{~A}]^{1 / 2}[\mathrm{C}]\), is caused by (a) doubling [A] (b) halving [C] (c) halving [A] (d) doubling [A] and [C]

For the reaction $$\mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{R}+\mathrm{Z} \quad \Delta H=+295 \mathrm{~kJ},$$ draw a reaction-energy diagram for the reaction if its activation energy is \(378 \mathrm{~kJ} .\)
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