Question

Assume that the following thermochemical reaction is known:

2A -> 3B; deltaH = -50 kJ
If 9.0 mol of B are formed during this reaction, what was the enthalpy change?

Answer 1

The enthalpy change for the reaction is -150 kJ.

Step-by-step explanation:

The enthalpy change can be determined by using the stoichiometry of the reaction. From the balanced equation 2A -> 3B, we can see that for every 2 moles of A, 3 moles of B are formed. Therefore, the ratio of moles of B to moles of A is 3/2. Given that 9.0 mol of B are formed, we can use this ratio to calculate the moles of A that reacted:

9.0 mol B x (2 mol A / 3 mol B) = 6.0 mol A

Since we know that for every 2 moles of A, the enthalpy change is -50 kJ, we can use this information to calculate the enthalpy change for 6.0 mol of A:

-50 kJ x (6.0 mol A / 2 mol A) = -150 kJ

Therefore, the enthalpy change for the reaction is -150 kJ.