Question

use hesss law and the equations shown below to determine the enthalpy of reaction for 2nocl(g)→n₂(g)+o₂(g)+cl₂(g) (1) ½n₂(g)+½o₂(g)→no(g) δh°=+90.5 kj (2) no(g)+½cl₂(g)→nocl(g) δh°=-39.0 kj (1)×-2: 2no(g)→n₂(g)+o₂(g) δh°=-181.0 kj (2)×-2: 2nocl(g)→2no(g)+cl₂(g) δh°=78.0 kj 2nocl(g)→n₂(g)+o₂(g)+cl₂(g) δhᵣ=(-181.0 kj)+(78.0 kj)=-103.0 kj

Explanation:

Step1: Reverse and scale equations

Reverse equation (1) and multiply by 2: \(2NO(g)
ightarrow N_{2}(g)+O_{2}(g)\), \(\Delta H = - 2\times90.5=-181.0\ kJ\)
Reverse equation (2) and multiply by 2: \(2NOCl(g)
ightarrow2NO(g)+Cl_{2}(g)\), \(\Delta H= - 2\times(-39.0) = 78.0\ kJ\)

Step2: Add the modified equations

Add the two modified equations:
\(2NO(g)
ightarrow N_{2}(g)+O_{2}(g)\) (\(\Delta H=-181.0\ kJ\))
\(2NOCl(g)
ightarrow2NO(g)+Cl_{2}(g)\) (\(\Delta H = 78.0\ kJ\))
Sum: \(2NOCl(g)
ightarrow N_{2}(g)+O_{2}(g)+Cl_{2}(g)\)

Step3: Calculate the enthalpy change

\(\Delta H_{r}=(-181.0\ kJ)+(78.0\ kJ)\)

Answer:

\(-103.0\ kJ\)