1) How many kJ are absorbed when 45.2 g of water at 31.3 oC is heated to 76.9 oC? 2) Calculate the total heat released in kcal when 72.1 g water at 25.2 oC is cooled to 0 oC and freezes. 3) How many kilojoules are required to heat 55,500 mg of gold with specific heat = 0.129 J/g oC is heated from 24.6 oC to 123.4 oC? 4) Calculate the heat needed in kcal to change 45.6 g of water at 100 oC to change into steam.

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

Final answer:

The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.

Explanation:

The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.

The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.

With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.

For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.

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Answer:

C

Explanation:

Answer:

Explanation:

Knowing that you have 64.92 grams of Hg(NO₃)₂ to make 5.00 liters of solution, you can calcualte the molarity of the solution.

Molarity is a measure of concentration, defined as the number of moles of solute per liter of soluiton. Mathematically:

Then, first you must calculate the number of moles of solute. The formula is:

You can either calculate the molar mass of the compound using the chemical formula or search it in the internet.

The molar mass of Hg(NO₃)₂  is found to be 324.7 g/mol.

Now you have everything to calculate the molarity of the solution:

Answer:

a. NaHCO₃ + HCl → NaCl + H₂O + CO₂

b. 39.14 g is the mass of NaHCO₃ required to produce 20.5 moles of CO₂

Explanation:

A possible reaction for NaHCO₃ to make dioxide is this one, when it reacts with hydrochloric to produce the mentioned gas.

NaHCO₃ + HCl → NaCl + H₂O + CO₂

Ratio in this reaction is 1:1

So 1 mol of baking soda, produce 1 mol of CO₂

Let's calculate the moles

20.5 g CO₂ / 44 g/m = 0.466 moles

This moles of gas came from the same moles of salt.

Molar mass baking soda = 84 g/m

Molar mass . moles = mass

84 g/m .  0.466 moles = 39.14 g

Answer:

Three atoms are attached to the central atom in NF3.

Explanation:

The central atom is always regarded as the atom having the least electronegativity in a molecule or ion. We can decide on what atom should be the central atom by comparing the relative electro negativities of the atoms in the molecule or ion.

If we consider NF3, we can easily see that nitrogen is less electronegative than fluorine, hence nitrogen is the central atom in the molecule. We can also observe from the molecular model that three atoms of fluorine were attached to the central atom. Hence there are three atoms attached to the central atom in the molecule NF3.

Answer:

4.93g are extracted

Explanation:

Partition coefficient (P) is defined as the ratio of solute dissolved in the organic solvent and the solute dissolved in the aqueous phase.

That is:

P = 7.5 = Concentration in dichloromethane / Concentration in water.

Knowing this, in the first extraction with 25mL of dichloromethane you will extract:

7.5 = (X/25mL) / (5g - X) / 100mL

Where X is the amount of compound A that is extracted.

7.5 = 100X / (125 - 25X)

937.5 - 187.5X = 100X

937.5 = 287.5X

3.26g of A are extracted in the first extraction.

In water will remain 5g - 3.26g = 1.74g

In the second extraction you will extract:

7.5 = (X/25mL) / (1.74g - X) / 100mL

7.5 = 100X / (43.5 - 25X)

326.25 - 187.5X = 100X

326.25 = 287.5X

1.13g are extracted in the second extraction.

And remain: 1.74g - 1.13g = 0.61g

In the third extraction you will extract:

7.5 = (X/25mL) / (0.61g - X) / 100mL

7.5 = 100X / (15.25 - 25X)

114.375 - 187.5X = 100X

114.375 = 287.5X

0.40g are extracted in the third extraction.

And remain: 0.61g - 0.40g = 0.21g

In the second extraction you will extract:

7.5 = (X/25mL) / (0.21g - X) / 100mL

7.5 = 100X / (5.25 - 25X)

39.375 - 187.5X = 100X

39.375 = 287.5X

0.14g are extracted in the fourth extraction.

Thus, after the three extractions you will extract: 0.14g + 0.40g + 1.13g + 3.26g = 4.93g are extracted

Final answer:

The process involves using the partitioncoefficient to determine how much of Compound A will prefer the dichloromethane solvent over the water. Following a calculation process through four rounds of extraction, it is concluded that approximately 4.999g of Compound A will be extracted using four 25mL portions of dichloromethane.

Explanation:

The partition coefficient of a compound is a measure of how much it prefers one solvent over another. Given that the partition coefficient of Compound A is 7.5 in dichloromethane with respect to water, we can predict how much of this compound could be extracted using four separate 25 mL portions of dichloromethane.

Here's the step-by-step calculation process:

1. We start with 5 grams of Compound A in 100 mL of water. Given the partition coefficient, in the initial phase, 5/(7.5+1)=0.625g remains in water and 7.5/8.5*5=4.375g goes into the dichloromethane.
2. After one extraction with 25ml of dichloromethane, the amount left in the water will be 0.625g*1/(7.5+1)=0.069g.
3. After the second extraction: 0.069g*1/(7.5+1) = 0.008g.
4. After the third extraction: 0.008g*1/(7.5+1) = 0.0009g.
5. After the fourth extraction: 0.0009g*1/(7.5+1) = 0.0001g.

In total, around 4.999g of compound A will be extracted using four 25mL portions of dichloromethane.

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