Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ