Lecture 19 Equilibrium II Worksheet 1)The equilibrium constant, K c , is 9.8 x 10 5 for H 2 (g) + S (s) H 2 S (g). a. Find the reaction quotient, Q c , if [H 2 ] = 0.762 M and [H 2 S] = 0.483 M . b. Has the process established equilibrium? If not, inwhich direction will it proceed? Justify your answer.

Q = 0.634

This means  we have more reactants than products. This will cause the reaction to shift to the right, the side of the products.

We have more H2 gas than H2S. The reaction will shift to the side of H2S

Explanation:

Step 1: Data given

Kc = 9.8 * 10^5

[H2]= 0.762 M

[H2S] = 0.483 M

When Q=K, the system is at equilibrium this means there will be no shift to  the left nor the right.

When Q<K, we have more reactants than products. This will cause the reaction to shift to the right, the side of the products.

When Q>K, we have more products than reactants. This will cause the reaction to shift to the leftt, the side of the reactants.

Step 2: The balanced equation

H2(g) + S(s) ⇆ H2S(g)

Step 3: Calculate Q

Q = [H2S] / [H2]

Q = (0.483 M) / (0.762 M)

Q = 0.634

Step 4: Calculate the direction

Q = 0.634

K = 9.8 * 10^5

Q <<<< K

This means  we have more reactants than products. This will cause the reaction to shift to the right, the side of the products.

We have more H2 gas than H2S. The reaction will shift to the side of H2S

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j