Question

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ 12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ Express your answer with the appropriate units.b. Calculate the enthalpy of the reaction4B(s)+3O2(g)→2B2O3(s)given the following pertinent information:B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJH2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJH2O(l)→H2O(g), ΔH∘D=+44 kJExpress your answer with the appropriate units.

Answer 1

(A)
`\Delta H^(\circ )_(r)= -144 kJ`

(B)
`\Delta H^(\circ )_(r)= - 2552kJ`

Step-by-step explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

`1/2 N_(2)(g)+O_(2)(g)\rightarrow NO_(2)(g), \Delta H^(\circ )_(a)=33.2 kJ....equation (a)`

`1/2N_(2)(g)+1/2O_(2)(g)\rightarrow NO(g), \Delta H^(\circ )_(b)=90.2 kJ ....equation (b)`

Now, multiplying equation (a) with 2:

⇒
`N_(2)(g)+2 O_(2)(g)\rightarrow 2 NO_(2)(g)....equation (a)`

Then equation b is reversed and multiplied with 2:

`2 NO(g)\rightarrow N_(2)(g)+ O_(2)(g)....equation (b)`

Now by adding the equation (a) and equation (b), we get:

⇒
`2 NO(g)+ \bcancel N_(2)(g)+\bcancel 2 O_(2)(g)\rightarrow 2 NO_(2)(g) +\bcancel N_(2)(g)+ \bcancel O_(2)(g)`

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

Therefore, the enthalpy of the reaction:

`\Delta H^(\circ )_(r)= 2* \Delta H^(\circ )_(a) - 2* \Delta H^(\circ )_(b)`

`= (2*33.2)- (2*90.2)=66.4 - 180.4= -144 kJ`

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

`B_(2)O_(3)(s)+3H_(2)O(g)\rightarrow 3O_(2)(g)+B_(2)H_(6)(g), \Delta H_(a )^(\circ )=+2035 kJ...equation (a)`

`2B(s)+3H_(2)(g)\rightarrow B_(2)H_(6)(g), \Delta H_(b )^(\circ )= +36 kJ...equation (b)`

`H_(2)(g)+1/2O_(2)(g)\rightarrow H_(2)O(l), \Delta H_(c )^(\circ )= -285 kJ...equation (c)`

`H_(2)O(l)\rightarrow H_(2)O(g), \Delta H_(d )^(\circ )=+44 kJ...equation (d)`

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

`6O_(2)(g)+2B_(2)H_(6)(g)\rightarrow 2B_(2)O_(3)(s)+6H_(2)O(g)...equation (a)`

`4B(s)+6H_(2)(g)\rightarrow 2B_(2)H_(6)(g)...equation (b)`

`6H_(2)O(l)\rightarrow 6H_(2)(g)+3O_(2)(g)...equation (c)`

`6H_(2)O(g)\rightarrow 6H_(2)O(l)...equation (d)`

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

Therefore, the enthalpy of the reaction:

`\Delta H^(\circ )_(r)= -2* \Delta H^(\circ )_(a) + 2* \Delta H^(\circ )_(b) - 6 * \Delta H_(c )^(\circ ) - 6 * \Delta H_(d )^(\circ )`

`= -2* (+2035 kJ)+ 2* (+36 kJ) - 6 * (-285 kJ)- 6 * (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ`