Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Stuck here, help me understand: - Chemical Thermodynamics - JEE Main

(\Delta H-\Delta U) for the formation of carbon monoxide (CO) from its elements at 298 K is

( R = 8.314 JK-1 mol-1  )

  • Option 1)

    –1238.78 J mol-1

  • Option 2)

    1238.78 J mol-1

  • Option 3)

    –2477.57 J mol-1

  • Option 4)

    2477.57 J mol-1

 

Answers (1)

best_answer

As we learnt in

Enthalpy of combustion -

Amount of Enthalpy change when 1 mole of substance is completly burn in excess of Oxygen

- wherein

CO_{(g)}+\frac{1}{2}O_{2}(g)\rightarrow CO_{2(g)}+283\ KJ

\Delta H_{comb} \: of\: CO= 283\, KJ

 

 C(S)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)

\Delta n_{g}=1-\frac{1}{2}=\frac{1}{2}

\Delta H-\Delta U=\Delta n_{g}RT=\frac{1}{2}\:J/mol\times8.314JK^{-1}mol^{-1}\times298

= 1238.78 J/mol


Option 1)

–1238.78 J mol-1

This option is incorrect.

Option 2)

1238.78 J mol-1

This option is correct.

Option 3)

–2477.57 J mol-1

This option is incorrect.

Option 4)

2477.57 J mol-1

This option is incorrect.

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission

Latest Question