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Pepsi [2]
2 years ago
11

The pressure at the center of the earth is probably greater than 3x106 atm, and the temperature there is about 4000 °C. What is

the change in the Gibbs function of reaction on going from crust to core for a reaction in which Vm = 1 cm3 mol-1 and Sm = 2.1 J K-1 mol-1
Chemistry
1 answer:
andrezito [222]2 years ago
7 0

Answer:

The change in the Gibbs function of reaction on going from crust to core, dG=291442.4 J/mol

Explanation:

dG=vdp-sdT

Where T is the temperature .

P is the pressure.

S is entropy

V= 1cm3 =10-6 m3

P= 3*10∧6 atm = 3*10∧11 pa

The temperature at the center of the earth, Tc = 4*103 °C. : The temperature at the sufrace (crust) of the earth, Ts =298K.

Subistuting the values

dG = (10-6 *(3*1011-105) - (2.1(4373-298))

dG=291442.4 J/mol.

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1. Using the balanced equation, answer the following questions:
nalin [4]

Answer:

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

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Answer:

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