Question

Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia:
N2(g) + 3H2(g) →2NH3(g)
ΔH=−92.kJ
In the second step, ammonia and oxygen react to form nitric oxide and water:
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g)
ΔH=−905.kJ
Calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions.

Answer 1

Answer: The net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen is -272.2 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reactions  are,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$     (1)

$\Delta H_1=-92kJ$

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$     (2)

$\Delta H_2=-905kJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$\frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)+\frac{5}{4}O_2(g)\rightarrow NO(g)+\frac{3}{2}H_2O$ $\Delta H_3=?$

According to the Hess’s law, if we multiply the reaction (1) by 1/2, (2) by 1/4 and then add (1) and (2)

So, the value $\Delta H_3$ for the reaction will be:

$\Delta H_3=\frac{1}{2}\times (-92kJ)+\frac{1}{4}\times (-905kJ)$

$\Delta H_3=-272.2kJ$

Hence the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen is -272.2 kJ

Answer 2

Answer:

ΔH = - 272.255 kJ      

Explanation:

Given that

In first step ,reaction is given as

N2(g) + 3H2(g) →2NH3(g)                  ΔH=−92 kJ

In second step ,reaction is given as

4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g)             ΔH=−905 kJ

Now multiple by in the first equation ,we get

2N2(g) + 6H2(g) →4NH3(g)                  ΔH=− 184 kJ

Now by adding the above equation

2N2(g) + 6H2(g) →4NH3(g)                  ΔH=− 184 kJ

4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g)             ΔH=−905 kJ

2 N2+6 H2 + 5 O2 → 4 NO + 6 H2O       ΔH=  - 1089 k J

For one mole of nitric oxide (NO)

$\Delta H=\dfrac{-1089}{4}= -272.255\ kJ$

ΔH = - 272.255 kJ

Therefore change in the enthalpy will be - 272.255 kJ.