Question

Calculate the enthalpy of the reaction 4b(s) + 3o2(g) → 2b2o3(s) given the following pertinent information: b2o3(s) + 3h2o(g) → 3o2(g), b2h6(g), Δh°a = 2035 kj; 2b(s) + 3h2(g) → b2h6(g), Δh°b = 36 kj; h2(g) + 1/2o2(g) → h2o(l), Δh°c = -285 kj; h2o(l) → h2o(g), Δh°d = 44 kj?

Answer 1

To find the enthalpy of the reaction 4B(s) + 3O2(g) → 2B2O3(s), the given reactions are reversed and manipulated according to Hess's Law, followed by summing their individual enthalpies to obtain the enthalpy change for the target reaction.

Step-by-step explanation:

To calculate the enthalpy of the reaction 4B(s) + 3O2(g) → 2B2O3(s) using Hess's Law, we will manipulate the given reactions and their enthalpies to match the target reaction. The standard enthalpies (ΔH°) for each reaction are given as:

• ΔH°a (B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g)) = 2035 kJ
• ΔH°b (2B(s) + 3H2(g) → B2H6(g)) = 36 kJ
• ΔH°c (H2(g) + 1/2O2(g) → H2O(l)) = -285 kJ
• ΔH°d (H2O(l) → H2O(g)) = 44 kJ

By reversing and manipulating these reactions, we can derive the enthalpy for our target reaction. First, we convert the formation of water from gas to liquid to align with ΔH° a and ΔH° b .

1. 3H2O(g) → 3H2O(l), ΔH = -3(44 kJ) because the reaction is the reverse of ΔH°d
2. The sum of ΔH°b and three times ΔH°c reversed gives us the formation of B2H6(g) from its elements.
3. Finally, we sum the reverse of ΔH°a with the enthalpies from steps 1 and 2 to obtain the enthalpy of the target reaction.

The calculations should be carried out with attention to stoichiometry, algebraic signs, and the conservation of matter to ensure that the resulting chemical equation is identical to the target reaction. By adding the enthalpies of the manipulated reactions together, we find the enthalpy of the target reaction.