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  • Please,please help me - Chemical Thermodynamics - JEE Main-2

If at 298 K the bond energies of C – H, C – C, C =  C and H – H bonds are respectively 414, 347, 615 and 435 kJ mol-1 , the value of enthalpy change for the reaction

H_{2}C=CH_{2(g)}+H_{2(g)}\rightarrow H_{3}C-CH_{3(g)}

at 298 K will be

  • Option 1)

    +250 kJ

  • Option 2)

    –250 kJ

  • Option 3)

    +125 kJ

  • Option 4)

    –125 kJ

 

Answers (1)

best_answer

As we learnt in

Heat of Reaction -

It is defined as the amount of  Enthalpy change in the reaction for given amount of reaction.

- wherein

C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}+393.5 KJ 

 

 CH_{2}=CH_{2}\:(g)+H_{2}\:(g)\rightarrow H_{3}C-CH_{3}\:(g)

4(C-H)=414x4=1656 6(C-H)=414x6=2484 1(H-H)=435x1=435
1(C=C)=615x1=615 1(C-C)3471x1347  
     

\Delta H\:of\ reactants =1656+615+435=2706

\Delta H\:of\ product =2484+347=2831

\therefore \Delta H\:of\ reaction=2706-2831=-125KJ 


Option 1)

+250 kJ

This option is incorrect 

Option 2)

–250 kJ

This option is incorrect 

Option 3)

+125 kJ

This option is incorrect 

Option 4)

–125 kJ

This option is correct 

Posted by

Aadil

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