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The standard reaction free energy AG=-64. kJ for this reaction: C(s) + 2Cl₂(g) →CC14(g) Use this information to complete the table below. Round each of your answers to the nearest kJ. reaction C(s) + C₁₂(g) → CC₁₂ (8) 2CC1 (g) → 2C(s) + 4C1₂ (8) CC14 (g) → C(s) + -> 2C1, (g) AG kJ KJ 0 X

The standard reaction free energy AG=-64. kJ for this reaction: C(s) + 2Cl₂(g) →CC14(g) Use this information to complete the table below. Round each of your answers to the nearest kJ. reaction C(s) + C₁₂(g) → CC₁₂ (8) 2CC1 (g) → 2C(s) + 4C1₂ (8) CC14 (g) → C(s) + -> 2C1, (g) AG kJ KJ 0 X

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The standard reaction free energy AG =-64. kJ for this reaction: C(s) + 2Cl₂(g) →CCl4(9) Use this information to complete the table below. Round each of your answers to the nearest kJ. reaction C(s) + = C₁₂(g) → 4 CC₁₂ (8) 2CC1 (g) 2C(s) + 4Cl₂ (g) CC1₂(g) → C(s) + 2Cl₂(g) AG KJ KJ KJ x10 Ś
Transcribed Image Text:The standard reaction free energy AG =-64. kJ for this reaction: C(s) + 2Cl₂(g) →CCl4(9) Use this information to complete the table below. Round each of your answers to the nearest kJ. reaction C(s) + = C₁₂(g) → 4 CC₁₂ (8) 2CC1 (g) 2C(s) + 4Cl₂ (g) CC1₂(g) → C(s) + 2Cl₂(g) AG KJ KJ KJ x10 Ś
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Step 1

 

The standard reaction free energy ∆°G= -64. kJ for this reaction:

C(s) +  2Cl2(g)  ------> CCl4(g)  ∆°G = -64 KJ  ......... A 

Divide the equation 'A' by 4 we get 

(1/4) C(s) +  (1/2) Cl2(g)  ------> (1/4) CCl4(g)  ∆°G = - 16 KJ  

 

Reverse the equation  'A' We get

CCl4(g) ------> C(s) +  2Cl2(g)  ∆°G = 64KJ 

Multiply the equation  'A' We get 

2 CCl4(g) ------> 2 C(s) +  4 Cl2(g)  ∆°G = 128 KJ  

 

 

 

 

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