Question

For the following reaction, 2HI(g) – H (9) +1,9 the equilibrium constant k, = 2.60X10-2 at 1110 K and 2.68x10-2 at 1140 K. Calculate AHⓇ for the reaction, assuming no change in AH between 1110 K and 1140 K. AH = kJ mol-1

Answer 1

10.63 kJ/mol is the
`\Delta H` of the reaction.

Step-by-step explanation:

To calculate
`\Delta H` of the reaction, we use Van't Hoff's equation, which is:

`\ln ((K_1)/(K_2))=(\Delta H)/(R)[(1)/(T_1)-(1)/(T_2)]`

where,

`K_(1)` = equilibrium constant at
`T_1`

`K_(2)` = equilibrium constant at
`T_2`

`\Delta H` = Enthalpy change of the reaction

R = Gas constant = 8.314 J/mol K

Given :
`K_1=2.60* 10^(-2)`

`K_2=2.68* 10^(-2)`

`T_1` = initial temperature =
`1110 K`

`T_2` = final temperature =
`1140 K`

`\Delta H` = ?

Putting values in above equation, we get:

`\ln((2.68* 10^(-2))/(2.60* 10^(-2)))=(\Delta H)/(8.314J/mol.K)[(1)/(1110 K)-(1)/(1140 K)]`

`\Delta H=10,627.617 J/mol=10.63 kJ/mol`

10.63 kJ/mol is the
`\Delta H` of the reaction.