Home
Class 12
CHEMISTRY
The enthalpy changes for the following p...

The enthalpy changes for the following process are listed below :
`Cl_(2)(g)=2Cl(g)," "242.3" kJ"mol^(-1)`
`I_(2)(g)=2I(g)," "151.0" kJ"mol^(-1)`
`ICl(g)=2I(g)+Cl(g)," "211.3" kJ"mol^(-1)`
`I_(2)(s)=I_(2)(g)," "62.76" kJ"mol^(-1)`
Given that standard states for iodine and chlorine are `I_(2)(s)` and `Cl_(2)(g)`, the standerd enthalpy of formation for ICl(g) is :

A

`-16.8 KJ mol^(-1)`

B

`+16.8 KJ mol^(-1)`

C

`+244.8 KJ mol^(-1)`

D

`-14.6 KJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(1)/(2) I_(2)(s) + (1)/(2) CI_(2)(g) rarr ICI(g)`
`DeltaH_(f,ICI) (g) = [(1)/(2)DeltaH_(I_(2))(s) rarr I_(2)(g) + (1)/(2)DeltaH_(I-I) + (1)/(2)Delta_(CI-CI)] - [DeltaH_(I-CI)]`
`=[(1)/(2)xx 62.76 + (1)/(2)151.0 + (1)/(2)xx 242.3] -[211.3]=16.73 KJ//mol`
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    RESONANCE|Exercise exercise-3 Part-3 :(Subjective questions)|6 Videos

  • THERMODYNAMICS

    RESONANCE|Exercise exercise-3 Part-3 :(Subjective questions)Section A|1 Videos

  • THERMODYNAMICS

    RESONANCE|Exercise exercise-3 Part-1 :(previous years)|8 Videos

  • TEST SERIES

    RESONANCE|Exercise CHEMISTRY|50 Videos

Similar Questions

Explore conceptually related problems

The enthaplpy changes state for the following processes are listed below: Cl_(2)(g)=2Cl(g) : 242.3KJmol^(-1) I_(2)(g)=2I(g) , 151.0KJ mol^(-1) ICl(g)=I(g)+Cl(g) : 211.3KJ mol^(-1) I_(2)(s)=l_(2)(g) , 62.76KJ mol^(-1) Given that the standard states for iodine chlorine are I_(2)(s) and Cl_(2)(g) , the standard enthalpy of formation for ICl(g) is:

The enthalpy change for the following process are listed below : (a) Cl_(2(g)) rarr 2Cl_((g)), DeltaH=242.3kJ mol^(-1) (b) I_(2(g))rarr 2I_((g)), DeltaH=151.0kJ mol^(-1) (c) ICl_((g)) rarr I_((g))+Cl_((g)), DeltaH=211.3kJ mol^(-1) (d) I_(2(s)) rarr I_(2(g)), DeltaH=62.76kJ mol^(-1) If standard state of iodine and chloride are I_(2(s)) and Cl_(2(g)) , the standard enthalpy of formation for ICl_((g)) is :

Using ._(f)G^(Theta)(Hi) = 1.3 kJ mol^(-1) , calculate the standard free enegry change for the following reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g)

Calculate the standard free energy change for the reaction : H_(2)(g) + I_(2)(g) to 2HI(g) DeltaH^(@) = 51.9 kJ "mol"^(-1) Given : S^(@)(H_(2)) = 130.6 J K^(-1) "mol"^(-1) S^(@)(I_(2)) = 116.7 J K^(-1) "mol"^(-1) and S^(@)(HI) = 206.3 J K^(-1) "mol"^(-1) .

Calculate Delta G (in kJ) for the reaction at 300 K, H_(2)(g)+Cl_(2)(g) rarr 2HCl (g) Given at 300 K, BE_(H-H) = 435 kJ mol^(-1), BE_(Cl-Cl) = 240 kJ mol^(-1), BE_(HCl) = 430 kJ mol^(-1) Entropies of H_(2), Cl_(2) and HCl are 131, 223 and 187 JK^(-1) mol^(-1) respectively.