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ziro4ka [17]
3 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]3 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]3 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 equilibrium for the indicator methyl orange is HC14H14SO3(red) + H2O C14H14SO3−(yellow) + H3O+ The reaction is exothermic.
muminat

Answer:

The solution will turn red.  

Explanation:

HC₁₄H₁₄SO₃ + H₂O ⇌ HC₁₄H₁₄SO₃⁻ +H₃O⁺

     (red)                           (yellow)

Methyl orange is a weak acid in which the ionized and unionized forms are distinct colours and are in equilibrium with each other,

At about pH 3.4,  the two the forms are present in equal amounts, and the indicator colour is orange.

If you add more acid, you are disturbing the equilibrium.

According to Le Châtelier's Principle, when you apply a stress to a system at equilibrium, it will respond in such a way as to relieve the stress.

The system will try to get rid of the added acid, so the position of equilibrium will move to the left.

More of the unionized molecules will form, so the solution will turn red.

 

4 0
3 years ago
For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

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