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ziro4ka [17]
4 years ago
7

½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole 21.0 kcal/mole + C(s) + 2S(s) → CS2(l) What type of reaction is represented by the

previous two examples?
Chemistry
2 answers:
wlad13 [49]4 years ago
7 0

<u>Answer:</u>

Exothermic Reaction are those reaction, in which energy is released while in endothermic reaction are those, in which energy is absorbed.

<u>Explanation:</u>

First Reaction:

As in this reaction, energy is released

½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole

so it is <em>exothermic reaction</em>

Second reaction:

As in this reaction, energy is absorbed

21.0 kcal/mole + C(s) + 2S(s) → CS2(l)

so it is <em>endothermic reactions</em>.


jeka57 [31]4 years ago
3 0

Answer:

Both the reactions are example of the endothermic reaction.

Explanation:

Endothermic reaction is a type of of chemical reaction in which energy is absorbed from the surrounding. The temperature of the surrounding decreases. The enthalpy of reaction is always positive \Delta H_{rxn}>0

\frac{1}{2}H_2(g) + \frac{1}{2}I_2(g)\rightarrow HI(g), \Delta H = +6.2 kcal/mol

The above reaction is an example of endothermic reaction.

21.0 kcal/mol + C(s) + 2S(s) \rightarrow CS_2(l)

This reaction is also an endothermic reaction in which energy is being added on the reactant side which indicates that energy is absorbed during the reaction.

Exothermic reaction is a type of of chemical reaction in which energy is released into the surrounding. The temperature of the surrounding increases..The enthalpy of reaction is always negative\Delta H_{rxn}

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The answer for the following problem is mentioned below.

  • <u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas (P_{1}) = 1.8 atm

Final pressure of the gas (P_{2})  = 4 atm

Initial temperature of the gas (T_{1}) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas (T_{2})

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

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(i.e.)

      <em> </em>\frac{P}{T}<em> = constant</em>

        \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1}  = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of a gas

        \frac{1.8}{4} = \frac{333}{T_{2} }

   T_{2} =\frac{333*4}{1.8}

    T_{2} = 740 K

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

8 0
3 years ago
Element x reacts with oxygen to produce x2o3 in an experiment it is found that 1.0000 g of x produces 1.1xxx g of x2o3 what is t
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Answer:

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

Explanation:

Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.

Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

Now, we have to balance Li by multiplying LiNO₂ by 2.

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The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst
OLEGan [10]

Answer:

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D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

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<em>     A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

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    <em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0
3 years ago
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The balanced chemical equation is :

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Thus clearly both reduction and oxidation are taking place.

Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.

To know more about oxidation number here:

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