For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol....

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Question

For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol. K). Calculate AG for this reaction at -19.0

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Answer 1

Answer #1

The problem can be solved using the relation

∆G = ∆H - T∆S

Where ∆G = Gibbs free energy

∆H = enthalpy change

∆S = entropy change.

First of all the value of ∆H for reaction at 239.1°C can be calculated. As this value will be independent of temperature, so the value of of ∆G at any other temperature can be calculated using same equation.

The problem is solved in the picture attached below.

AG= 363.40 kilmol 363400 Il mol OS:739-34J/molk T 239.1°C = 239.1 + 273 - 512.ik OG DH TAS 363400 DH - 512-1 x 739.34 363400

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Answer 2

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For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol....

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For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol....

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For a particular reaction at 203.3 °C, AG = 424.99 kJ/mol, and AS - 551.38 J/mol...

For a particular reaction at 212.3 °C, AG = -889.99 kJ/mol, and AS = 385.90 J/(mol · K). Calculate AG for this reaction...

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For a particular reaction at 156.7 °C, AG = 279.62 kJ, and AS = 253.74 J/K....

For a particular reaction at 205.1205.1 °C, ΔG=−45.90 kJ/molΔG=−45.90 kJ/mol, and ΔS=395.61 J/(mol⋅K)ΔS=395.61 J/(mol⋅K). Calculate ΔG...

For a particular reaction at 172.4 °C, Δ?=−269.73 kJ/mol ΔG=−269.73 kJ/mol, and Δ?=970.33 J/(mol⋅K) Calculate ΔG...

For a particular reaction at 124.8 °C, Δ?=−841.80 kJ/mol , and Δ?=646.52 J/(mol⋅K) . Calculate ΔG...

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For a particular reaction at 227.0C, Δ?=−1398.89 kJ/mol, and Δ?=300.75 J/(mol⋅K) Calculate ΔG for this reaction at...

For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol....

Homework Answers

Add Answer to: For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol....

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Add Answer to:
For a particular reaction at 239.1 °C, AG = 363.40 kJ/mol, and AS = 739.34 J/mol....