Question

Consider the following equations:

3A+6B3D ∆H-401 kJ
E+2F A ∆H-103.3 kJ
C→E+3D ∆H = 65.7 kJ
Suppose the first equation is reversed and multiplied by 1/6, the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction?

Answer 1

To find the net reaction and the overall enthalpy change, we adjust the given equations by reversing, multiplying, or dividing as directed, and then apply the principles of Hess's law. We sum the adjusted enthalpy changes accordingly to determine the net ΔH for the resultant combined equation.

Step-by-step explanation:

We are tasked to find the net reaction by manipulating given chemical equations using Hess's law. To begin, we need to apply two principles where (1) if a chemical reaction is reversed, the sign on ΔH is changed, and (2) if a multiple of a chemical reaction is taken, the same multiple of the ΔH is taken as well.

Let's consider the following adjusted equations based on the directions provided:

1. Reverse the first equation and multiply by 1/6:
2. Divide the second equation by 2:
3. Divide the third equation by 2:

By adding these adjusted equations together, we can eliminate terms that appear on both sides of the resultant equation. This way, we will arrive at the net reaction. During this process, remember to adjust the enthalpy changes (ΔH) accordingly, based on the alterations made to each equation, then sum them alongside the combined chemical equations.

It's important to note that according to Hess's law, the ΔH for a reaction that is the sum of two or more reactions is the sum of the ΔH values for the individual reactions. Hence, we directly apply this law to our calculations to determine the overall enthalpy change for the net reaction. Now, we can combine the modified equations, taking into account the reversal or scaling that has been applied, and sum the enthalpy values to find the net ΔH.