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.

(A)

(B)

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

Now, multiplying equation (a) with 2:

⇒

Then equation b is reversed and multiplied with 2:

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

⇒  

⇒  2NO(g) + O₂(g) → 2NO₂(g)

Therefore, the enthalpy of the reaction:

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)    

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

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

4B(s)+3O₂(g) → 2B₂O₃(s)

Therefore, the enthalpy of the reaction: