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MrMuchimi
3 years ago
6

A considerable amount of heat is required for the decomposition of aluminum oxide. 2 al2o3(s) â 4 al(s) + 3 o2(g) δh = 3352 kj

(a) what is the heat change for the formation of 1 mol of aluminum oxide from its elements?
Chemistry
1 answer:
wolverine [178]3 years ago
5 0
To determine the heat change during a reaction, we need to know the heat of reaction of the process. For this case, the heat of formation is needed since it is a formation reaction. It will have units of kJ per mole. We determine the heat change as follows:

Heat change = 3352 kJ / 2 mol Al2O3( 1 mol Al2O3) = 1676 kJ
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0,02 g/mL = 20 g/L

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8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?​
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Answer:

the frequency of photons v = 1.509\times10^{39}Hz

Explanation:

Given:  first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

\Delta E_1 = 1000KJ/mol

We know that plank's constant

h = 6.626\times10^{-34} Js

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

= \frac{10^6J}{6.626\times10^{-34} Js}

v = 1.509\times10^{39}Hz

Hence, the frequency of photons v = 1.509\times10^{39}Hz

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3 years ago
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr
lys-0071 [83]

Answer:

a. No change.    

b. The equilibrium will shift to the right.

c. No change

d. No change

e.  The equilibrium will shift to the left

f.  The equilibrium will shift to the right      

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining   Keq  again.

The equilibrium constant  for  A(s)⇌B(g)+C(g)  

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use  Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a.  double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´     ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp  ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to  double the pressures of A and B. Therefore there is no change.

b.  double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change  reduced the pressures by a half :

Q =  pB´x pC´  = (  1/2 pB ) x ( 1/2 pC )  =  1/4 pB x pC  = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the  concentration of B and halve the concentration of C

Doubling the concentrantion doubles  the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q =  pB´x pC´  = ( 2 pB ) x ( 1/2 pC ) = K

e.  double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q =  pB´x pC´   = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f.  double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q =  pB´x pC´   = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16  Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

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Answer:

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Explanation:

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