Question

Reaction 1: c3h8(g)+5o2(g)→3co2(g)+4h2o(g), δh1=−2043 kj reaction 2: 6c3h8(g)+30o2(g)→18co2(g)+24h2o(g), δh2=?

Answer 1

Answer : The value of
`\Delta H_2` for the reaction is -12258 kJ/mole.

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,

`C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)`

`\Delta H_1=-2043KJ`

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

`6C_3H_8(g)+30O_2(g)\rightarrow 18CO_2(g)+24H_2O(g)`
`\Delta H_2=?`

According to the Hess’s law, if we multiply the reaction by 6 then the
`\Delta H` will also multiply by 6.

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

`\Delta H_2=6* (-2043kJ/mole)`

`\Delta H_2=-12258kJ/mole`

Hence, the value of
`\Delta H_2` for the reaction is -12258 kJ/mole.

Answer 2

Answer :
`\Delta H_2=12258KJ`

Explanation :

The given reaction 1 :

`C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)`

`\Delta H_1=2043KJ`

The given reaction 2 :

`6C_3H_8(g)+30O_2(g)\rightarrow 18CO_2(g)+24H_2O(g)`

`\Delta H_2=?`

The reaction 2 is the 6 times of the reaction 1. Therefore, the value for
`\Delta H_2` is also 6 times of
`\Delta H_1`.

`\Delta H_2=6* \Delta H_1=6* 2043KJ=12258KJ`

Therefore,
`\Delta H_2=12258KJ`