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aliina [53]
3 years ago
12

For the equilibrium reaction N2O4(g) ⇀↽ 2 NO2(g) taking place inside a sealed container fitted with a piston, predict the effect

of raising the piston (to increase the volume of the container) on the composition of the mixture. 1. The equilibrium will shift to favor formation of N2O4(g). 2. There will be no effect on the composition. 3. More information is needed. 4. The equilibrium will shift to favor formation of NO2(g).
Physics
1 answer:
Butoxors [25]3 years ago
6 0

Answer:

4. The equilibrium will shift to favor formation of NO2(g)

Explanation:

According to La Chatalier's Principle which states that when an equilibrium system undergoes changes either in temperature, volume or concentration; there will be in a change in the system in order to reach equilibrium.

From the above equation,

N2O4(g) ⇀↽ 2 NO2(g)

From the above reaction, there are 2 moles of gaseous product on the left and 1 mole of gaseous reactant.

Therefore, there are more moles of gases in the left hand side than the right hand side.

Because a decrease in volume favors the direction that produces fewer moles, an increase in volume will therefore shift this system towards the side with more moles of gases that is, more products are formed hence, this system will shift to right and produce more moles of products i.e more NO2(g) formed.

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The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

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x = 6 cm = 0.06 m

k = 400 N

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F = 6N

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Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

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Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

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½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

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