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Nadusha1986 [10]
3 years ago
15

Hydroxyl radicals react with and eliminate many atmospheric pollutants. However, the hydroxyl radical does not clean up everythi

ng. For example, chlorofluorocarbons - which destroy stratospheric ozone - are not attacked by the hydroxyl radical. Consider the hypothetical reaction by which the hydroxyl radical might react with a chlorofluorocarbon: OH(g)+CF2Cl2(g)-->HOF(g)+CFCl2(g)Use bond energies to explain why this reaction is improbable. Calculate \Delta Hrxn of this reaction.
Chemistry
1 answer:
Nesterboy [21]3 years ago
7 0

Answer:

ΔHreaction = 263.15 kJ/mol

Explanation:

The reaction is as follow:

OH + CF₂Cl₂ → HOF + CFCl₂

You need to calculate the enthalpy of reaction and for this it is necessary to know the standard enthalpies for each of the compounds. These enthalpies are as follows and can be found in your textbook or on the Internet.

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

delta(H)_{reaction} =((1*(-97.8)+(1*(-92))-((1*39)+(1*(-491.15))=263.15kJ/mol

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What source of thermal energy does an internal combustion engine use?
Dafna1 [17]

Answer:

D

Explanation:

The source of the thermal energy in a combustion engine is the spark from the spark plug that ignites the fuel and air mixture converting chemical energy into thermal energy.

5 0
3 years ago
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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
Which order is correct in listing the bond lengths from shortest to longest? A. single, double, triple B. triple, double, single
Alex787 [66]
A I believe would be the answer
3 0
3 years ago
Calculate the amount of energy required to melt 500 g of ice at 0oc. (δhfus=5.96 kj/mole; δhfus is the energy required to melt i
marta [7]
When you want to melt an ice, you only need the latent energy of fusion, <span>δhfus. We use the given value, then multiply this with the given amount to determine the amount of energy. Since the energy is per mole basis, use the molar mass of ice which is 18 g/mol. The solution is as follows:

</span>ΔH = 5.96 kJ/mol * 1 mol/18 g * 500 g
<em>ΔH = 165.56 kJ</em><span>
</span>
3 0
3 years ago
A solution contains the ions ag+ , pb2+ and ni2+. dilute solutions of nacl, na2so4 and na2s are available to separate the positi
sladkih [1.3K]
 <span>Use the sequence E (NaCl, Na2SO4, then Na2S). Silver is insoluble as a chloride, so it would be removed first, the others (Pb and Ni) are soluble as chlorides(Note; lead chloride is soluble as a hot solution but will ppt when cold), next, PbSO4 is insoluble but NiSO4 is soluble so use Na2SO4 to separate lead from nickel. Lastly, nickel sulfide is insoluble and can be separated and collected.

Hope I helped :)</span>
6 0
3 years ago
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