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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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1 mole NH3 -------------- 17.031 g
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=> 20.31 moles of NH3

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Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

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Given there is 100 g of nitrogen and 100 g of hydrogen

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Number of moles of hydrogen = 100 ÷ 2 = 50

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From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

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∴ Limiting reactant is nitrogen and excess reactant is hydrogen

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