Question

What is the change in enthalpy for the following reaction?

4NH3(g) + 302(g) 2N2(g) + 6H2O(l)

Given:
NH3: ∆H= -46.2 kJ
Given: H2O: ∆H= -286 kJ

Answer 1

Answer:

Explanation:

multiplying reaction (1) by 2 .. 2H2 + O2 ----> 2H2O delta H = -242 X 2 = -484 adding this to reaction (2) 2H2 + O2 + N2 + 2H2O -------> 2H2O + N2H4 + O2 ...... delta H = -484 + 534 = 50 O2 and 2H2O will cancel out from both sides ... 2H2 + N2 -----> N2H4 .... delta H = 50 kj/mole

Answer 2

Hey mate...

here's the answer...

2 N2 + 6 H2 = 4 NH3 , ΔH = (1/-46)^4

6 H2O = 6 H2 + 3 O2 , ΔH = (1/484)^3

-------- ------------ ----------- -------------- +

2N2 + 6H2O = 3O2 + 4NH3 , ΔH = (1/-46)^4 x (1/484)^3 = 1.97 x 10^-15

Hope this helps❤