Whats the process of solving this?

Question

A chemist measures the enthalpy change ΔH during the following reaction:

2Na+Cl2→2NaCl ΔH=-822KJ

Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.

NaCl→Na+1/2Cl2

4Na+2Cl2→4NaCl

2NaCl→2Na+Cl2

Answer 1

When you reverse a reaction, you simply change the sign for ∆H

If you multiply a reaction by 2, you multiply the ∆H by two, etc.

2Na + Cl₂ ==> 2NaCl ... ∆H = -822 kJ

NaCl ==> Na + 1/2 Cl₂ This is reversed and halved. So the ∆H needs to be halved and change the sign

∆H = + 1/2 x 822 = +411 kJ

4Na + 2Cl₂ ==> 4NaCl This is the same reaction only doubled. So the ∆H keeps the same sign but has to be doubled. ∆H = 2 x -822 = - 1644 kJ

2NaCl ==> 2Na + Cl₂ This is the original reaction in reverse. So, we simply just change the sign.

∆H = +822 kJ