Question

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ?Given:PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = +157 kJP4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ

Answer 1

Answer:

Therefore  $\bigtriangledown H^\circ_{rxn}$= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       $\bigtriangledown H^\circ_{rxn}$=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       $\bigtriangledown H^\circ_{rxn}$= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     $\bigtriangledown H^\circ_{rxn}$= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          $\bigtriangledown H^\circ_{rxn}$= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          $\bigtriangledown H^\circ_{rxn}$=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    $\bigtriangledown H^\circ_{rxn}$=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      $\bigtriangledown H^\circ_{rxn}$= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       $\bigtriangledown H^\circ_{rxn}$= -1835 KJ

Therefore  $\bigtriangledown H^\circ_{rxn}$= -1835 KJ