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Neko [114]
3 years ago
13

PLZ HELP For the reaction: 2NO2(g) → N2O4(l),

Chemistry
1 answer:
Furkat [3]3 years ago
6 0

Answer: \Delta H_{rxn}=-20kJ/mol-(+66kJ/mol)

Explanation:

Heat of reaction or enthalpy change is the energy released or absorbed during the course of the reaction.

It is calculated by subtracting the enthalpy of reactants from the enthalpy of products.

\Delta H=H_{products}-H_{reactants}

\Delta H = enthalpy change = ?

H_{products} = enthalpy of products

H_{reactants} = enthalpy of reactants

For the given reaction :

2NO_2(g)\rightarrow N_2O_4(l)

\Delta H=H_{N_2O_4}-2\times H_{NO_2}

\Delta H=-20kJ/mol-(+66kJ/mol)

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Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react
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Answer:

ΔH  = - 272 kJ

Explanation:

We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:

N2(g) +         3H2(g) →          2NH3(g)                 ΔH=−92.kJ  Multiplying by 2:      

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__________________________________________________

2N2(g) +   6H2(g) + 5O2(g)→  4NO(g)  + 6H2O(g)      ΔH = (-184 +(-905 )) kJ

                                                                                     ΔH =    -1089 kJ

Notice how the intermediate NH3 cancels out.

As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH  for the formation  of one mol NO:

-1089 kJ/4 mol NO  x 1 mol NO =  -272 kJ (rounded to nearest kJ)

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