Lawrie Ryan and Roger Norris Solutions for Chapter: Entropy and Gibbs Free Energy, Exercise 6: Question

Author:Lawrie Ryan & Roger Norris

Lawrie Ryan Chemistry Solutions for Exercise - Lawrie Ryan and Roger Norris Solutions for Chapter: Entropy and Gibbs Free Energy, Exercise 6: Question

Attempt the free practice questions on Chapter 23: Entropy and Gibbs Free Energy, Exercise 6: Question with hints and solutions to strengthen your understanding. Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years) solutions are prepared by Experienced Embibe Experts.

Questions from Lawrie Ryan and Roger Norris Solutions for Chapter: Entropy and Gibbs Free Energy, Exercise 6: Question with Hints & Solutions

EASY
AS and A Level
IMPORTANT

Calculate the standard entropy change of the system in the given below reaction using the standard molar entropy values given here. (Values for Sin J K-1 mol-1).

Given:

H2O2l=109.6H2Ol=69.90O2g=205.0

2H2O2l2H2Ol+O2g

EASY
AS and A Level
IMPORTANT

Calculate the standard entropy change of the system in the given below reaction using the standard molar entropy values given here. (Values for Sin J K-1 mol-1).

Given:

NH4NO3s=151.1N2Og=219.7H2Og=69.90NH4NO3sN2Og+2H2Og

EASY
AS and A Level
IMPORTANT

Calculate the standard entropy change of the system in the given below reaction using the standard molar entropy values given here. (Values for Sin J K-1 mol-1).

Given:

MgOs=26.90O2g=205.0Mgs=32.702Mgs+O2g2MgOs

EASY
AS and A Level
IMPORTANT

Calculate the standard entropy change of the system in the given below reaction using the standard molar entropy values given here. (Values for 

Sin J K-1 mol-1).

Given:

NaCls=72.10Cl2g=165.0Nas=51.202Nas+Cl2g2NaCls

EASY
AS and A Level
IMPORTANT

Calculate the standard entropy change of the system in the given below reaction using the standard molar entropy values given here. (Values for  

Sin J K-1 mol-1).

Given:

Fes=27.30MgOs=26.90Mg(s)=32.70Fe2O3s=87.403Mg(s)+Fe2O3s3MgOs+2Fes