Acetylene gas, C2H2( g ), is used in welding. Part A Part complete Write a balan

Question

Acetylene gas, C2H2(g), is used in welding.

Part A

Part complete

Write a balanced equation for the combustion of acetylene gas to CO2(g) and H2O(l).

Express your answer as a chemical equation. Identify all of the phases in your answer.

2C2H2(g)+5O2(g)?4CO2(g)+2H2O(l)

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Correct

Part B

How much heat is produced in burning 3 mol of C2H2 under standard conditions if both reactants and products are brought to 298 K?

Express your answer using five significant figures.

nothing

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Incorrect; Try Again; 5 attempts remaining

Part C

What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction when 3 mol of C2H2 is burned?

Express your answer using five significant figures.

nothing

Acetylene gas, C2H2(g), is used in welding.

Part A

Part complete

Write a balanced equation for the combustion of acetylene gas to CO2(g) and H2O(l).

Express your answer as a chemical equation. Identify all of the phases in your answer.

2C2H2(g)+5O2(g)?4CO2(g)+2H2O(l)

SubmitPrevious Answers

Correct

Part B

How much heat is produced in burning 3 mol of C2H2 under standard conditions if both reactants and products are brought to 298 K?

Express your answer using five significant figures.

?H =

nothing

  kJ  

SubmitPrevious AnswersRequest Answer

Incorrect; Try Again; 5 attempts remaining

Part C

What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction when 3 mol of C2H2 is burned?

Express your answer using five significant figures.

wmax =

nothing

  kJ  

Explanation / Answer

part B

2C2H2(g)+5O2(g) -----> 4CO2(g)+2H2O(l)

DHrxn = (4*DH0f,CO2(g)+2*DH0f,H2O(l))-(2*DH0f,C2H2(g)+5*DH0f,O2(g))

      = (4*-393.5+2*-285.8)-(2*226.7+5*0)

      = -2599 kj

2 mol C2H2 = -2599 KJ

3 mol C2H2 = -2599*3/2 = -3898.5 Kj

answer: 3.8985*10^3 kj

PART c

dGrxn = (4*DG0f,CO2(g)+2*DG0f,H2O(l))-(2*DG0f,C2H2(g)+5*DG0f,O2(g))

      = (4*-394.4+2*-237.1)-(2*209.2+5*0)

      = -2470.2 kj

dGrxn = Wmax

from equation,

maximium work(Wmax) per 2 molC2H2 = -2470.2 kj

   3 mol C2H2 = -2470.2*3/2 = -3705.3 kj
  
answer: -3.7053*10^3 kj

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