# Are these calculations valid enough? [closed]

I am not able to understand how these calculations are being done in this book. Up until this section, the book is going smoothly. However, how are these calculations accountable? If anyone can explain this to me then it will be valid enough for me.

• What's the book? What part exactly do you think is flawed and why? – andselisk Apr 21 '19 at 6:49
• BOOK NAME : INTRODUCTION TO THE THERMODYNAMICS OF MATERIALS How are the moles calculated and what is the exact basis of the addition of 1 mole for calculations? – Oumang Kaushik Apr 21 '19 at 6:54
• You should include this into your question and add proper bibliographical reference including not just the name of the book, but the authors/editors, edition, publisher, year and ISBN. Besides, it's not moles that are calculated, it's amount of substance. – andselisk Apr 21 '19 at 7:00
• I am learning sir, from the next time it will be even more specific for all of us. – Oumang Kaushik Apr 21 '19 at 7:07
• I'm not criticizing your question, I'm just making it easier for others to quicker understand what it's about; and there is no need to call me sir, that's internet after all:) – andselisk Apr 21 '19 at 7:22

Equations (iii) through (v) are based on the relation

$$\Delta G = \Delta H - T \Delta S$$

which holds as long as $$T$$ is constant (which it is in the problem). The values of $$\Delta H$$ and $$\Delta S$$ have been looked up and substituted, likely in tables provided in the book.

The next four (unnumbered) equations are based on the relationship

$$K_p = \exp \left( \frac{-\Delta G}{R T} \right)$$

where $$K_p$$ is the equilibrium constant and $$\Delta G$$ is a function of $$T$$ evaluated using the previous relationships.

The fractions discussed in the text relate to the stoichiometry. For example, consider the first ratio:

• $$K_p = 19.1$$, so at equilibrium there are 19.1 H$$_2$$O molecules for every H$$_2$$ molecule
• Assuming that the initial state before the reaction had no H$$_2$$O, all of the H atoms which end up in H$$_2$$O molecules must have originally been in the form of H$$_2$$. The balanced chemical equation shows that producing a H$$_2$$O molecule requires exactly one H$$_2$$ molecule, so for every H$$_2$$ molecule in the final state there were 20.1 in the initial state
• Since 19.1/20.1 of the H$$_2$$ molecules react an 1/20.1 do not, and since one molecule of CoO is consumed per molecule of H$$_2$$ consumed, it follows that each mole of H$$_2$$ provided can react with 19.1/20.1 = 0.95 moles of CoO.
• Thanks for your valuable time and attention to this question. @user1476176 – Oumang Kaushik Apr 21 '19 at 15:02