Question

The enthalpy change for the reaction of titanium metal with gaseous iodine is given by the following thermochemical equation: 2 Ti(s) + 3 12(g) rightarrow 2 Til3(s) delta H = -839 kJ What is the enthalpy change for the reaction below? Til3(s) rightarrow Ti(s) + 3/2 12(g)

A) -1.68 x 102 kJ
B) +4.20 x 102 kJ
C) -4.20 x 102 kJ
D) +8.39 x 102 kJ

Answer 1

Answer:
`4.20* 10^2kJ`

Step-by-step explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

`2Ti(s)+3I_2(g)\rightarrow 2TiI_3(g)`

`\Delta H_1=-839KJ`

Now we have to determine the value of
`\Delta H` for the following reaction i.e,

`TiI_3(g)\rightarrow Ti(s)+\farc{3}{2}I_2(g)`

`\Delta H_2=?`

According to the Hess’s law, if we reverse the reaction , the sign of enthalpy changes

`2TiI_3(g)\rightarrow 2Ti(s)+3I_2(g)`
`\Delta H_3=+839KJ`

According to the Hess’s law, if we half the reaction , the enthalpy changes to half

`TiI_3(g)\rightarrow Ti(s)+(3)/(2)I_2(g)`
`\Delta H_2=(+839)/(2)KJ`

So, the value
`\Delta H_2` for the reaction will be:

`\Delta H_2=419.5kJ`

`\Delta H_2=4.20* 10^2kJ`

Hence, the value of
`\Delta H_2` for the reaction is
`4.20* 10^2kJ`