Hess's Law Problem

Calculate the reaction enthalpy for the synthesis of hydrogen chloride gas:

H2(g) + Cl2(g) ---> 2HCl(g)

Given that:

NH3(g) + HCl(g) ---> NH4Cl(s), Delta H = -176.0 kJ
N2(g) + 3H2(g) ---> 2NH3(g), Delta H = -92.22 kJ
N2(g) + 4H2(g) + Cl2(g) ---> 2NH4Cl(s), Delta H = -628.86 kJ

You're trying to get the top equation from the three you have below.

I'll just change the equations around so that they fit together (you can multiply, divide and reverse reactions to change them, but remember to also change the delta H if you do so)

2NH4Cl(s) ---> 2NH3(g) + 2HCl(g), Delta H = +352kJ (i flipped the sign to positive and multiplied by 2)
2NH3(g) ---> N2(g) + 3H2(g) , Delta H = +92.22 kJ (I reversed the reaction and flipped the sign)
N2(g) + 4H2(g) + Cl2(g) ---> 2NH4Cl(s), Delta H = -628.86 kJ

We can now combine the three equations together can cancel any terms that are on both sides.
This cancels all of the 2NH4Cl (s), N2 and 2NH3 so we're left with:

H2(g) + Cl2(g) ---> 2HCl(g)

We just need to add the three delta H values: 352 + 92.22 - 628.86 = -184.64kJ

It's pretty easy. I hope this helped

the given reactions are :

NH3(g) + HCl(g) ---> NH4Cl(s)........(1) Delta H = -176.0 kJ
N2(g) + 3H2(g) ---> 2NH3(g)..........(2), Delta H = -92.22 kJ
N2(g) + 4H2(g) + Cl2(g) ---> 2NH4Cl(s).........(3), Delta H = -628.86 kJ

we have to calculate the enthalpy of reaction [H2(g) + Cl2(g) ---> 2HCl(g)]
reversing the reaction no. 1 and multiplying with 2
2NH4Cl(s) ---> 2NH3(g) + 2HCl(g)...........Delta H = 176.0 kJ.........(4)

reversing the reaction no 2
2NH3(g) ----> N2(g) + 3H2(g)...........Delta H = 92.22 kJ...........(5)

now adding the reaction (3) , (4) & (5)

N2(g) + 4H2(g) + Cl2(g) ---> 2NH4Cl(s).........., Delta H = -628.86 kJ

2NH4Cl(s) ---> 2NH3(g) + 2HCl(g)...........Delta H = 176.0 kJ

2NH3(g) ----> N2(g) + 3H2(g)...........Delta H = 92.22 kJ
______________________________________…
we get H2(g) + Cl2(g) ---> 2HCl(g).........Delta H = [-628.86+176.0+92.22] = -360.64 kJ
{after eliminating N2, NH4Cl, NH3}

Hence, the enthalpy of the reaction H2(g) + Cl2(g) ---> 2HCl(g) is -360.64 kJ......