Consider the following reaction:

H2(g) + I2(s) ---> 2HI(g)
TriangleH = 51.9 kj (enthalpy)

Calculate the standard enthalpies of reaction for the following reaction:

6HI(g) ---> 3H2(g) + 3I2(s)

Thanks!

You want the reverse of 3x the rxn.

That will be -51.9*3 kJ

To calculate the standard enthalpies of reaction for the given reaction, we need to use the given enthalpy change for the original reaction and apply it to the stoichiometry of the target reaction.

Given reaction:
H2(g) + I2(s) ---> 2HI(g) ΔH = -51.9 kJ

Target reaction:
6HI(g) ---> 3H2(g) + 3I2(s)

The approach to solving this is as follows:

1. Multiply the given reaction by 3 so that the number of HI molecules matches in both reactions.

3(H2(g) + I2(s) ---> 2HI(g)) ΔH = -51.9 kJ

2. Reverse the direction of the given reaction because we want to calculate the enthalpies of the reverse reaction.

-3(H2(g) + I2(s) ---> 2HI(g)) ΔH = +51.9 kJ

3. Add the multiplied and reversed reaction to the target reaction to create the overall reaction:

6HI(g) - 3(H2(g) + I2(s)) ---> 3H2(g) + 3I2(s) ΔH(total) = ΔH(1) + ΔH(2)

So, the overall reaction is:

3HI(g) ---> 1/2H2(g) + 1/2I2(s) ΔH(total) = 3(-51.9 kJ) + (+51.9 kJ)

Since the ΔH for the given reaction is -51.9 kJ, the standard enthalpy change for the target reaction is:

ΔH(total) = 3(-51.9 kJ) + (+51.9 kJ) = -155.7 kJ

Therefore, the standard enthalpy of reaction for the target reaction is -155.7 kJ.