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ryzh [129]
3 years ago
6

The reaction: 2 CO(l) + O2(g) ⇄2 CO2(g) with a H = −25.0 kJ/mol, is at equilibrium. Which of the following lists three ways thi

s reaction can be shifted to the right?A. remove CO2, lower pressure and remove COB. remove CO2, raise pressure and add COC. raise temperature, lower pressure and remove O2D. add O2, raise pressure and lower temperatureE. remove CO2, increase volume and lower temperature
Chemistry
1 answer:
elena-s [515]3 years ago
8 0

Answer:

Option B and D both have 3 ways to shift the reaction to the right

Explanation:

The reaction  2 CO(l) + O2(g) ⇄2 CO2(g) is exothermic because ΔH = -25 kJ < 0 This means heat will be released

Therefore, increasing the temperature will shift the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the right.

A. remove CO2, lower pressure and remove CO

⇒ Lowering the pressure will shift the equilibrium to the side with most moles of gas : On the left side there are 3 moles, on the right side 2 moles. By lowering the pressure, the equilibrium will shift to the<u> left.</u>

B  remove CO2, raise pressure and add CO

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the <u>right</u> side.

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The<u> right</u> side).

⇒ Add CO: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added CO. (The<u> right</u> side).

C raise temperature, lower pressure and remove O2

⇒ Increasing the temperature will shift the equilibrium to the <u>left</u>

D add O2, raise pressure and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the <u>right</u>

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the <u>right</u> side.

⇒ Add O2: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added O2. (The<u> right</u> side).

E  remove CO2, increase volume and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the <u>right</u>

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The <u>right</u> side).

⇒ Increase volume : with a pressure decrease due to an increase in volume, the side with more moles is more favorable. The equilibrium will shift to the <u>left</u> side.

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3 years ago
A large balloon contains 5400 m3 of He gas that is kept at a temperature of 280 K and an absolute pressure of 1.10 x 105 Pa. Fin
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Answer:

1.02 × 10⁶ g

Explanation:

Step 1: Given data

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  • Temperature (T): 280 K
  • Absolute pressure (P): 1.10 × 10⁵ Pa
  • Molar mass of He (M): 4.002 g/mol

Step 2: Convert "V" to L

We will use the conversion factor 1 m³ = 1000 L.

5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 101325 Pa.

1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm

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We will use the ideal gas equation.

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n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K

n = 2.56 × 10⁵ mol

Step 5: Calculate the mass of He (m)

We will use the following expression.

m = n × M

m = 2.56 × 10⁵ mol × 4.002 g/mol

m = 1.02 × 10⁶ g

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