Question

calculate $delta h^{circ}$ for the following reaction:

$4 mathrm{nh}_{3}(g)+5 mathrm{o}_{2}(g) \to 4 mathrm{no}(g)+6 mathrm{h}_{2} mathrm{o}(g)$

from the following information:

$operatorname{rxn}(mathrm{i}) mathrm{n}_{2}(g)+mathrm{o}_{2}(g) \to 2 mathrm{no}(g) quad delta h=-180 mathrm{~kj}$
$operatorname{rxn}(mathrm{ii}) mathrm{n}_{2}(g)+3 mathrm{h}_{2}(g) \to 2 mathrm{nh}_{3}(g) quad delta h=-92 mathrm{~kj}$
$operatorname{rxn}(mathrm{iii}) 2 mathrm{h}_{2}(g)+mathrm{o}_{2}(g) \to 2 mathrm{h}_{2} mathrm{o}(g) quad delta h=-484 mathrm{~kj}$

describe how $operatorname{rxn}(mathrm{i})$ should be manipulated.

after $operatorname{rxn}(mathrm{i})$ is manipulated, what is the resulting value for $delta h$?

Explanation:

Step1: Analyze the target reaction and rxn(i)

The target reaction has \(4NO(g)\) as a product. Rxn(i) has \(2NO(g)\) as a product. So, rxn(i) should be multiplied by \(2\).

Step2: Calculate the new \(\Delta H\) for the manipulated rxn(i)

If a reaction is multiplied by a factor \(n\), its \(\Delta H\) is also multiplied by \(n\). For rxn(i): \(\Delta H=- 180\space kJ\). After multiplying by \(2\), \(\Delta H = 2\times(-180)\space kJ\)

Answer:

• \( - 360\)