Question

- Calculate the Standard Enthalpy of the reaction below:

NH3(g) + HCI (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCI(g) → H2(g) + Cl2(g)
AH = +184.6 KJ
2H2(g) + 1/2 N2(g) + 1/2 Cl2(g) → NH4Cl(s) deltaH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
deltaH = +184.6 kJ​

Answer 1

Step-by-step explanation:

We have the three equations:

`NH_(3(g)) + HCl_((g)) => NH_4Cl_((s)) ..... \Delta H = ? (1)\\2HCl_((g)) => H_(2(g)) + Cl_(2(g)) .... \Delta H = +184.6 kJ (2)\\2H_(2(g)) + 1/2N_(2(g)) + 1/2Cl_(2(g)) => NH_4Cl_((s)) ..... \Delta H = -314.2 kJ (3)\\ N_(2(g)) + 3H_(2(g)) => 2NH_(3(g)) .... \Delta H = +184.6kJ (4)`

(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.