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jok3333 [9.3K]
3 years ago
6

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

_______.
Chemistry
1 answer:
nata0808 [166]3 years ago
8 0

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g).

When \rm H_2 is added to \rm H_2C\text{=}CH_2 (ethene) under heat and with the presence of a catalyst, \rm H_3C\text{-}CH3 (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, \Delta S < 0.

The equation for the change in Gibbs Free Energy for a particular reaction is:

\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}}).

For a particular reaction, the more negative \Delta G is, the more spontaneous ("favorable") the reaction would be.

Since typically \Delta S < 0 for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

T (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

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To demonstrate the formation of iron (iii) chloride from iron fillings
Pavel [41]

Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.

<h3>How can iron (iii) chloride be formed from iron fillings?</h3>

Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:

  • Iron fillings + Cl₂ → FeCl₃

Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.

In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.

Learn more about iron (iii) chloride at: brainly.com/question/14653649

#SPJ1

5 0
2 years ago
If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial
jeka94

Answer:

the  initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

KSCN \rightarrow K^ + SCN^-

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5  m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the  initial concentration of SCN- in the mixture is 0.00588 M

5 0
3 years ago
(a) the temperature on a warm summer day is 87 °f. what is the temperature in °c? (b) many scientific data are reported at 25
olchik [2.2K]

Answers:

              (a)  30.55 °C

              (b) 298 K and 77°F

              (c)  204.44 °C and 477.44 K

              (d)  -320.8 °F and -196 °C

Explanation:

Converting °C into °F;

                                   °F  =  °C × 1.8 + 32

Converting °F into °C;

                                   °C  =  °F - 32 ÷ 1,8

Converting °C into K;

                                   K  =  °C + 273

Converting K into °C;

                                   °C  =  K - 273

8 0
3 years ago
If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of
navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

M=\frac{n}{V}

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

2.60g*\frac{1mol}{102.89g} =0.0253mol

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

M=\frac{0.0253mol}{0.160L}=0.158M

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L

That in milliliters turns out to be:

V=0.211L*\frac{1000mL}{1L}=211mL

Learn more about molarity: brainly.com/question/10053901

6 0
2 years ago
Answer pls pls plss this is my mock
Serggg [28]

Answer:

if this is a mock then just guess

Explanation:

4 0
3 years ago
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