Question

Determine delta sub r H in Kj/mol for this reaction. 4No+6H2O --> 4NH3 +5O2 using the equations and the enthalpy change of the reactions given.

N2 + O2 --> 2NO delta sub r H= 180.1

NH3 --> 1/2 N2 +1.5H2. delta sub r H= 54.3

2H2O --> 2H2 + O2 delta sub r H= 486.7

Answer 1

$Given,\; N_{2}+O_{2}\rightarrow 2NO\; \; \; \; \Delta H=180.1KJ/mol$

$4NO\rightarrow 2N_{2}+2O_{2}\; \; \; \; \Delta H=-180.1*2=-360.2KJ/mol\Rightarrow equation\; 1$

$Given,\; NH_{3}\rightarrow \frac{1}{2}N_{2}+1.5H_{2}\; \; \; \; \Delta H = 54.3KJ/mol\Rightarrow equation\; 2$

$2N_{2}+6H_{2}\rightarrow 4NH_{3}\; \; \; \; \Delta H =-54.3*4=-217.2KJ/mol$

Given, 2H2O + 2H2 + O2 AH = 486.7KJ/mol

$6H_{2}O\rightarrow 6H_{2}+3O_{2}\; \; \; \; \Delta H=486.7*3=1460.1KJ/mol\Rightarrow equation\; 3$

$4NO+6H_{2}O\rightarrow 4NH_{3}+5O_{2}\; \; \; \Delta H=(-360.2-217.2+1460.1)KJ/mol$

$Therefore,\; \Delta H\; of\; this\; reaction\; is\; \; 882.7KJ/mol$