Question

QUESTION 1. The following data is provided for use in answering the following questions: Substance | ?H°f (kJmol?¹ at 25 °C) | S° (Jmol?¹K?¹ at 25 °C) NO (g) | 91.3 | 210.8 O2 (g) | 0.00 | 205.2 NO2 (g) | 33.2 | 240.1 Consider the reaction between NO (g) and O2 (g) to give NO2 (g) at a constant pressure of 1 atm and a temperature of 25 °C. 2 NO (g) + O2 (g) ? 2 NO2 (g) a) Calculate the value of ?G°rxn for the above reaction under these conditions and indicate whether it is spontaneous or not. b) Calculate the value of the equilibrium constant for the reaction if it is brought to equilibrium under these conditions. c) Calculate the value of the new equilibrium constant if the reaction temperature is raised from 25 °C to 250 °C.

          QUESTION

1. The following data is provided for use in answering the following questions:

Substance | ?H°f (kJmol?¹ at 25 °C) | S° (Jmol?¹K?¹ at 25 °C)
NO (g) | 91.3 | 210.8
O2 (g) | 0.00 | 205.2
NO2 (g) | 33.2 | 240.1

Consider the reaction between NO (g) and O2 (g) to give NO2 (g) at a constant pressure of 1 atm and a temperature of 25 °C.

2 NO (g) + O2 (g) ? 2 NO2 (g)

a) Calculate the value of ?G°rxn for the above reaction under these conditions and indicate whether it is spontaneous or not.

b) Calculate the value of the equilibrium constant for the reaction if it is brought to equilibrium under these conditions.

c) Calculate the value of the new equilibrium constant if the reaction temperature is raised from 25 °C to 250 °C.

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QUESTION

1. The following data is provided for use in answering the following questions:

Substance | ?H°f (kJmol?¹ at 25 °C) | S° (Jmol?¹K?¹ at 25 °C)
NO (g) | 91.3 | 210.8
O2 (g) | 0.00 | 205.2
NO2 (g) | 33.2 | 240.1

Consider the reaction between NO (g) and O2 (g) to give NO2 (g) at a constant pressure of 1 atm and a temperature of 25 °C.

2 NO (g) + O2 (g) ? 2 NO2 (g)

a) Calculate the value of ?G°rxn for the above reaction under these conditions and indicate whether it is spontaneous or not.

b) Calculate the value of the equilibrium constant for the reaction if it is brought to equilibrium under these conditions.

c) Calculate the value of the new equilibrium constant if the reaction temperature is raised from 25 °C to 250 °C.

Added by Manuela W.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Pravin N

02/20/2022

Step 1

2 kJ/mol ΔG°f(NO) = 210.8 kJ/mol ΔG°f(O2) = 205.2 kJ/mol ΔG°rxn = 2(33.2) - [2(210.8) + 205.2] ΔG°rxn = 66.4 - (421.6 + 205.2) ΔG°rxn = 66.4 - 626.8 ΔG°rxn = -560.4 kJ/mol Show more…

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The following data is provided for use in answering the following questions: Substance | ΔH°f (kJmol⁻¹ at 25 °C) | S° (Jmol⁻¹K⁻¹ at 25 °C) NO (g) | 91.3 | 210.8 O2 (g) | 0.00 | 205.2 NO2 (g) | 33.2 | 240.1 Consider the reaction between NO (g) and O2 (g) to give NO2 (g) at a constant pressure of 1 atm and a temperature of 25 °C. 2 NO (g) + O2 (g) → 2 NO2 (g) a) Calculate the value of ΔG°rxn for the above reaction under these conditions and indicate whether it is spontaneous or not. b) Calculate the value of the equilibrium constant for the reaction if it is brought to equilibrium under these conditions. c) Calculate the value of the new equilibrium constant if the reaction temperature is raised from 25 °C to 250 °C.

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Transcript

0:00 Hello students.

00:01 Today we will discuss about the calculation of free energy change and the equilibrium constant.

00:08 In this question, this equation is given.

00:12 That is no and oxygen reacts to from no2.

00:17 In this question, no and oxygen are the reactants and no2 is the product.

00:26 The equation is happening on a temperature of 25 degrees centigrade, that means to 98 degree kelvin.

00:35 And for each substances, the enthalpy change and the entropy changes are given.

00:42 To calculate the delgy, that is the free energy change, we first have to calculate the whole enthalpy change and the whole entropy change by using this two formulas.

00:55 For the enthalpy change, we have to first calculate.

01:00 The total enthalpy for the products, then we have to subtract from it the total enthalpy change of the reactor.

01:11 Similarly, for the entropy change, first we have to know the entropy of the product, then we have to subtract the entropy of the reactor.

01:24 So, let's start.

01:29 The del age for this reaction will be, first we have to react.

01:34 Down the product that is 33 .2.

01:37 So we have to subtract 91 .3 plus 0.

01:44 From this we will get minus 81 .1 kilojou per million.

01:54 For this questions we have to notice that here kilojoule is given and here jule is given so we have to convert everything in one unit so we are converting this to jule and we are getting 58 -1 -0 jule per moon now we will calculate the entropy change that is 240 .1 minus 2 10 .8 plus 200 this will give us a value of minus 175 .9 jule per mole per kelvin...

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