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2 years ago

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For which of the following reactions is ΔHrxn equal to ΔHf of the product? You do not need to look up any values to answer this

question. Select all that apply. Group of answer choices 1/2 O2(g) + H2O(g) LaTeX: \longrightarrow ⟶ H2O2(g) Na+(g) + F-(g) LaTeX: \longrightarrow ⟶ NaF(s) K(g) + 1/2 Cl2(g) LaTeX: \longrightarrow ⟶ KCl(s) O2(g) + 2 N2(g) LaTeX: \longrightarrow ⟶ 2 N2O(g) None of the above

1 answer:

Ipatiy [6.2K]2 years ago

7 0

Answer:

In none of the reactions ΔH°rxn equal to ΔH°f of the product.

Explanation:

The standard enthalpy of formation (ΔH°f) is the enthalpy change when 1 mole of a product is formed from its constituent elements in the standard states.

1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)

ΔH°rxn is NOT equal to ΔH°f of the product because H₂O(g) is not an element but a compound.

Na⁺(g) + F⁻(g) ⟶ NaF(s)

ΔH°rxn is NOT equal to ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).

K(g) + 1/2 Cl₂(g) ⟶ KCl(s)

ΔH°rxn is NOT equal to ΔH°f of the product because K is not in its standard state (K(s)).

O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)

ΔH°rxn is NOT equal to ΔH°f of the product because 2 moles of N₂O are formed.

In none of the above ΔHrxn equal to ΔHf of the product.

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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

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2 years ago

If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration

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2 years ago

C2H5OH(aq) + MnO− 4 (aq) → Mn2+(aq) + CH3COOH(aq) of acetic acid from ethanol by the action of permanganate ion in acidic soluti

Andreas93 [3]

Answer :

Ethanol $(C_2H_5OH)$ act as reducing agent.

The smallest possible integer coefficient of $MnO_4^-$ in the combined balanced equation is, 4

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given chemical reaction is,

$C_2H_5OH(aq)+MnO_4^-(aq)\rightarrow CH_3COOH(aq)+Mn^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $C_2H_6O\rightarrow C_2H_4O_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}$

Now balance oxygen atom on both side.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}+4H_2O$

Now balance hydrogen atom on both side.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^+$

Reduction : $MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O$

Now balance the charge.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^++4e^-$

Reduction : $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 4 and then added both equation, we get the balanced redox reaction.

Oxidation : $5C_2H_6O+5H_2O\rightarrow 5C_2H_4O_2+20H^++20e^-$

Reduction : $4MnO_4^-+32H^++20e^-\rightarrow 4Mn^{2+}+16H_2O$

The balanced chemical equation in acidic medium will be,

$5C_2H_6O+4MnO_4^-+12H^+\rightarrow 5C_2H_4O_2+4Mn^{2+}+11H_2O$

In the redox reaction ethanol act as reducing agent and permanganate ion act as an oxidizing agent.

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2 years ago

Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C

timurjin [86]

Answer:

The specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-$

Reduction half reaction: $2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-$

Thus, the specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

8 0

3 years ago