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2 years ago

Consider this reaction: NH4 HPO42− → NH3 H2PO4− Which is the Bronsted-Lowry acid ? NH4 HPO42- HPO42– H2PO4–.

Chemistry

1 answer:

kvv77 [185]2 years ago

8 0

Bronsted - Lowry acid in the given reaction is NH₄, as it gives H⁺ ion.

<h3>What is Bronsted - Lowry acid?</h3>

According to the theory of Bronsted - Lowry, acids are those substances which gives H⁺ ion or proton in the aqueous medium.

Given chemical reaction is :

NH₄ + HPO₄²⁻ → NH₃ + H₂PO₄⁻

In the above reaction NH₄ is the Bronsted - Lowry acid as it gives H⁺ ion in the reaction and changes to NH₃ which is the conjugate base of NH₄. Whereas HPO₄²⁻ is the Bronsted - Lowry base as it accepts the H⁺ ion to form H₂PO₄⁻ which is the conjugate acid of it.

Hence, option (1) is correct, i.e. NH₄ is the Bronsted - Lowry acid.

To know more about Bronsted - Lowry acid, visit the below link:

brainly.com/question/1435076

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Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

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ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

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The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

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