Chapter 20, Problem 70AE

(a)

Interpretation Introduction

Interpretation: The catalyst in the below mentioned scheme of the carbon-nitrogen cycle that occurs in the sun, needs to be determined.

  

Concept Introduction: The carbon-nitrogen cycle which occurs in the sun is a fusion reaction, wherein Hydrogen gets converted into Helium.

Answer to Problem 70AE

The catalyst in the mentioned scheme of carbon-nitrogen cycle is $\text{C}$ .

Explanation of Solution

Following is the carbon-nitrogen cycle that occurs in the sun:-

  $\begin{array}{l}H+C\to N+\gamma \\ N\to C+e\\ H+C\to N+\gamma \\ H+N\to O+\gamma \\ O\to N+e\\ H+N\to C+He+\gamma \end{array}$

The overall reaction is $4H\to He+2e$

A catalyst is a substance that influences the speed of a chemical reaction, while at the same time remains unchanged. In the mentioned scheme of carbon-nitrogen cycle that occurs in the sun, $\text{C}$ is the catalyst, as the same is not consumed in the reaction, but only initiates the reaction.

(b)

Interpretation Introduction

Interpretation: The nucleons that are intermediates in the below mentioned scheme of the carbon-nitrogen cycle that occurs in the sun needs to be determined.

  

Concept Introduction: The carbon-nitrogen cycle which occurs in the sun is a fusion reaction, wherein Hydrogen gets converted into Helium.

Answer to Problem 70AE

The nucleons that are intermediates in the mentioned scheme are $\text{N}\text{, }\text{C}\text{, }\text{N}\text{, }\text{O}\text{ and }\text{N}$ .

Explanation of Solution

Following is the carbon-nitrogen cycle that occurs in the sun:-

  $\begin{array}{l}H+C\to N+\gamma \\ N\to C+e\\ H+C\to N+\gamma \\ H+N\to O+\gamma \\ O\to N+e\\ H+N\to C+He+\gamma \end{array}$

The overall reaction is $4H\to He+2e$

The molecular entity which is evolved from the reactants and will further react in order to give the products, in a chemical reaction, are called as intermediates. In this case, the nucleon intermediates formed are $\text{N}\text{, }\text{C}\text{, }\text{N}\text{, }\text{O}\text{ and }\text{N}$ .

(c)

Interpretation Introduction

Interpretation: The energy that is released per mole of Hydrogen nuclei in the below mentioned scheme of the carbon-nitrogen cycle that occurs in the sun needs to be determined.

  

Concept Introduction: The carbon-nitrogen cycle which occurs in the sun is a fusion reaction, wherein Hydrogen gets converted into Helium.

Answer to Problem 70AE

  $-5.95\times {10}^{11}J/gH$ Is the energy released per mole of Hydrogen nuclei?

Explanation of Solution

In order to start with, the mass defect $\left(\Delta m\right)$ for $\text{H}\text{e}$ needs to be calculated. The quantity of energy that gets released can be calculated by comparing the sum of the masses of neutrons and protons with the mass of the Helium nucleus.

In the Helium atom $\text{H}\text{e}$ the number of protons = atomic number = 2

The number of neutrons = $\text{mass number - atomic number}$

  $\begin{array}{l}=4-2\\ =2\end{array}$

The number of electrons = atomic number = 2

The overall reaction is as shown below:-

  $4H\to He+2e$

  $\begin{array}{l}\text{Mass defect }\left(\Delta m\right)=\left[\text{Mass of }\text{H}\text{e nucleus+2}\left(\text{mass of positron}\right)\right]\\ -\left[4\left(\left(\text{mass of }\text{H}\text{ nucleus}\right)-\left(\text{mass of electron}\right)\right)\right]\\ =\left[\begin{array}{l}\left(\text{atomic mass of }\text{H}\text{e}-2\left(\text{mass of electron}\right)\right)\\ +2\left(\text{mass of positron}\right)\end{array}\right]\\ -\left[4\left(\left(\text{mass of }\text{H}\text{ nucleus}\right)-\left(\text{mass of electron}\right)\right)\right]\\ =\left[\left(4.00260u\right)\right]-\left[4\left(\left(1.00727u\right)-\left(5.486\times {10}^{-4}u\right)\right)\right]\\ =\left[\left(4.00260u\right)\right]-\left[4\left(1.00727u\right)\right]\\ =-0.02649u\end{array}$

The above derived is the difference in the mass per nucleus.

Further the energy change $\left(\Delta E\right)$ can be calculated using the following equation:-

  $\Delta E=\Delta m{c}^2$

Wherein

  $\begin{array}{l}\left(\Delta m\right)=\left(-0.02649\times \frac{u}{nucleus}\right)\times \left(1.66\times {10}^{-27}\frac{kg}{u}\right)\\ =-4.40\times {10}^{-29}Kg/nucleus\end{array}$

Speed of light $\left(c\right)$ = $3.00\times {10}^{8}m/s$

Substituting these values in the energy change equation:-

  $\begin{array}{l}\Delta E=\Delta m{c}^2\\ =\left(-4.40\times {10}^{-29}kg/nucleus\right)\times {\left(3.00\times {10}^{8}m/s\right)}^2\\ =-3.96\times {10}^{-12}J/nucleus\text{ because 1J=1Kg }{m}^2/s^2\end{array}$

Hence $3.96\times {10}^{-12}J$ of energy is released per nucleus formed and the negative sign is an indication that the process was exothermic. Therefore, the energy released for the formation of 1 mole of $\text{H}\text{e}$ is given as follows:-

  $\left(-3.96\times {10}^{-12}J/nucleus\right)\times \left(6.022\times {10}^{23}nuclei/mol\right)=-2.38\times {10}^{12}J/mol$

Further the energy released per mole of $\text{H}$ will be

  $\begin{array}{l}=2.38\times {10}^{12}\frac{J}{molHe}\times \frac{1molHe}{4molH}\\ =-5.95\times {10}^{11}J/gH\end{array}$

Therefore the energy released per mole of Hydrogen nuclei is $-5.95\times {10}^{11}J/gH$