CHEMISTRY COLLEGE

A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −86.00 J/K · mol. Determine the temperature (in °C) below which the reaction is spontaneous.

Answers

Answer 1

Answer:

Answer:

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Explanation:

Step 1: Data given

ΔH = −320.1 kJ/mol

ΔS = −86.00 J/K · mol.

Step 2: Calculate the temperature

ΔG<0 = spontaneous

ΔG= ΔH - TΔS

ΔH - TΔS <0

-320100 - T*(-86) <0

-320100 +86T < 0

-320100 < -86T

320100/86 > T

3722.1 > T

The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)

We can prove this with Temperature T = 3730 K

-320100 -3730*(-86) <0

-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous

T = 3700 K

-320100 -3700*(-86) <0

-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous

The temperature is quite high because of the big difference between ΔH and ΔS.

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Related Questions

ΔH is always positive for a ...........a. endothermic reactionb. exothermic reactionc. spontaneous reaction.d. nonspontaneous reaction.
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50.0ml each of 1.0M Hcl and 1.0M Naoh at room temperature (20.0c) are mixed the temperature of the resulting Nacl solutions increase to 27.5cthe density if the resulting Nacl solutuion 1.02 g/ml
the specific heat of the resulting Nacl solutions is 4.06j/gc
calculate the heat of neutralisation of hcl and naoh in kj/mol nacl products

Answers

Answer:

62.12kJ/mol

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

You can find the released heat of the reaction and heat of neutralization (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be finded with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), thus:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that are reacting releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

62.12kJ/mol is heat of neutralization

Which of the following is an oxidation-reduction reaction? Which of the following is an oxidation-reduction reaction? Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) Pb(C2H3O2)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaC2H3O2(aq) All of the above are oxidation-reduction reactions.

Answers

Answer:

NONE OF THE ABOVE

Explanation:

None of the above are examples of an oxidation - reduction or a redox reaction . This is because there is no change in the oxidation state of any of the elements in the reaction when the reaction happens .

For a redox reaction , transfer of electrons must take place.
The oxidation states must increase or decrease in atleast 2 of the elements of a compound.
For example :

$Zn^(2+)+Cu$ ⇒ $Cu^(2+)+Zn$

What are two functions of the cilia?

Answers

Answer:

- Proper urine flow by signalling the kidney cells.

- They act as mechanoreceptors or sensory receptors.

Explanation:

Motile' (or moving) cilia are found in the lungs, respiratory tract and middle ear. These cilia have a rhythmic waving or beating motion. They work, for instance, to keep the airways clear of mucus and dirt, allowing us to breathe easily and without irritation. They also help propel sperm.
Btw I don’t know if this is what you meant

If Nurse Antonio adds that 7 grams of NaCI to water to make 1 liter of solution, what is the molar concentration of the solution? Use dimensional analysis and show how you completed the unit conversion.

Answers

Given :

Nurse Antonio adds that 7 grams of NaCI to water to make 1 liter of solution.

To Find :

The molar concentration of the solution .

Solution :

Molecular mass of NaCl , $m=58.44\ g/mol$ .

Now , number of moles is given by :

$n=\frac{\text{Given weight}}{\text{Molecular Mass}}\n\nn=(7)/(58.44)\ mole\n\nn=0.12\ moles$

Molarity is given by :

$M=\frac{\text{Number of moles}}{\text{Volume ( in Liters)}}\n\nM=(0.12)/(1)\ M\n\nM=0.12\ M$

Hence , this is the required solution .

Claims • Evidence • Reasoning Makea claim about ways people can stay safe
during storms with high wind and heavy
rains. Summarize evidence to support the
claim and explain your reasoning.

Answers

Answer:

the claim is that when people avoid storms they hide in any secret place in there house

Explanation:

when people do that they don't even have to worry about a single thing that will happen to them

A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 358C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.

Answers

Answer:

(a) $\Delta S_(ref)=3.876(kJ)/(K)$

(b) $S_(heat\ source)=-1.678(kJ)/(K)$

Explanation:

Hello,

(a) In this case, such refrigerant, we can notice that at the given conditions, the initial entropy from property tables (Cengel 7th ed) is:

$s_(initial)=s_f+xs_(fg)=0.15457+0.4*0.78316=0.4678(kJ)/(kg*K)$

Now, for the final condition, we first need to compute the initial specific volume as it remains the same (rigid tank) after the thermodynamic process:

$v_(initial)=v_f+xv_(fg)=0.0007533+0.4*(0.099867-0.0007533)=0.0404(m^3)/(kg)$

Then, at 400 kPa we evaluate the given volume that is also between the liquid and vapor specific volume, thus, we calculate the quality at the end of the process:

$x_f=(0.0404-0.0007907)/(0.051201-0.0007907) =0.786$

With it, we compute the final entropy:

$s_(final)=0.24761+0.785*0.67929=0.781(kJ)/(kg*K)$

Finally, entropy change for the refrigerant turns out:

$m_(ref)=(0.5m^3)/(0.0404(m^3)/(kg) )=12.4kg \n\n\Delta S_(ref)=12.4kg *(0.781(kJ)/(kg*K)-0.4678(kJ)/(kg*K) )\n\n\Delta S_(ref)=3.876(kJ)/(K)$

(b) In this case, by using the first law of thermodynamics we compute the acquired heat by the refrigerant from the heat source by computing the initial and final internal energy respectively (no work is done):

$Q=\Delta U$

$u_(initial)=38.28+0.4*186.21=112.764(kJ)/(kg)\n \nu_(final)=63.62+0.786*171.45=198.40(kJ)/(kg)$

Hence:

$Q=12.4kg*(198.40-112.764)(kJ)/(kg) =1059.1kJ$

Finally, the entropy change of the heat source (which release the heat, therefore it is negative):

$S_(heat\ source)=(1059.1kJ )/((358+273.15)K) \n\nS_(heat\ source)=-1.678(kJ)/(K)$

$\Delta S _(tot)=3.876(kJ)/(K)-1.678(kJ)/(K)\n\n\Delta S _(tot)=2.198(kJ)/(K)$

Which has thermodynamic agreement as it is positive

Regards.

Final answer:

The entropy changes in this process can be partially calculated using principles from thermodynamics. However, without the exact heat transfer, not all values can be determined.

Explanation:

The calculation of the entropy change in this thermodynamic process involves principles from thermodynamics and requires steps to determine the initial and final states of the refrigerant. First, we would need to find the entropy at the initial and final states using the refrigerant properties table for refrigerant-134a and the provided information (200 kPa and 40% quality initially, 400 kPa finally). The entropy change of the refrigerant is the difference between the final and initial entropy.

Next, the entropy change of the heat source is calculated as the heat transfer divided by the absolute temperature of the source. However, the problem does not provide the amount of heat transferred from the source, making it impossible to determine this value directly.

Finally, in an isolated system, the total entropy change of the process is the sum of the entropy changes of the refrigerant and the heat source. Here, the precise values cannot be calculated due to a lack of specific data including exact heat transfer.

Learn more about Entropy Change here:

brainly.com/question/35154911

#SPJ3

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