Question

Consider the following intermediate chemical equations.

What is the enthalpy of the overall chemical reaction

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Answer 1

The enthalpy change for the overall reaction is 999 kJ.

To find the enthalpy change for the overall reaction
`PCl_(5)`(g) →
`PCl_(3)`(g) +
`Cl_(2)`(g), we need to combine the given intermediate equations.

The first equation is:
`P_(4)`(s) + 6
`Cl_(2)`(g) → 4
`PCl_(3)`(g), with an enthalpy change of Δ
`H_(1)` = -2,439 kJ.

The second equation is: 4
`PCl_(6)`(g) →
`P_(4)`(s) + 10
`Cl_(2)`(g), with an enthalpy change of Δ
`H_(2)` = 3,438 kJ.

By reversing the first equation and multiplying the second equation by 4, we can cancel out
`P_(4)` (s) and
`Cl_(2)` (g) to obtain the overall reaction:
`PCl_(5)`(g) →
`PCl_(3)`(g) +
`Cl_(2)`(g).

Since enthalpy change is a state function, we can sum up the enthalpy changes of the intermediate reactions to calculate the enthalpy change for the overall reaction:

Δ
`H_(overall)` = -2,439 kJ + 3,438 kJ = 999 kJ

Answer 2

Answer: 250 kJ

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 Hess’s law, the chemical equation can be treated as algebraic expressions 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.

`P_4(s)+6Cl_2\rightarrow 4PCl_3`
`\Delta H_1=-2439kJ` (1)

`4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)`
`\Delta H_2=3438kJ` (2)

Net chemical equation:

`PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)`
`\Delta H=?` (3)

Adding 1 and 2 we get,

`4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2`
`\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ` (4)

Now dividing equation (4) by 4, we get

`PCl_5(g)\rightarrow PCl_3(g)+Cl_2`

`\Delta H=(\Delta H_3)/(4)=(1000kJ)/(4)=250kJ` (4)