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3 years ago

Be able to describe the relationship between Freedom of Movement and phase change.

Chemistry

1 answer:

lisabon 2012 [21]3 years ago

7 0

Ion knoe Wdym by “be able to describ’ so ima put it in my own words idr lol:)

if you talm bout some kentic energy or sum ok but other Dan dat ion knoe tbh

I can explain how transferring kinetic energy in and out of a substance can cause a change

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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

In a 51.0-g aqueous solution of methanol ch4o the mole fraction of methanol is 0.270. what is the mass of each component?

gladu [14]

Mole fraction is another way of expressing the concentration of a solution or mixture. It is equal to the moles of one component divided by the total moles.

mass CH4O = 0.270 (51) = 13.77 g methanol

water = 51 - 13.77 = 37.23 g water

Hope this answers the question. Have a nice day.

4 0

3 years ago

What have chemists done to help pople conserve energy

Ainat [17]

Chemists have developed insulation.

4 0

3 years ago

Arrange in terms of increasing atomic radius. From smallest to largest. S, F, Sr, Na, Ga

BaLLatris [955]

Answer:

F<S<Na<Ga<Sr

Explanation:

Atomic radius increased from the right of the periodic table to left, and from the top to the down

5 0

3 years ago

The US mail delivery considered a network true or false

Svetllana [295]

True
Hope this helps!!❤️

6 0

3 years ago