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3 years ago

Which two changes would make this reaction reactant-favored?

Chemistry

1 answer:

Vera_Pavlovna [14]3 years ago

4 0

Two changes would make this reaction reactant-favored

C. Increasing the temperature

D. Reducing the pressure

<h3>Further explanation</h3>

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

The formation of H₂O is an exothermic reaction (releases heat)

If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient

in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2

As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater

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umka21 [38]

Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.

Further Explanation:

The Ideal Gas Equation is:

$PV = nRT$

where:

P - pressure (in atm)

V - volume (in L)

n - amount of gas (in moles)

R - universal gas constant $0.08206 \frac{L-atm}{mol-K}$

T - temperature (in K)

In the problem, we are given the values:

P = 2.00 atm (3 significant figures)

V = 3.00 L (3 significant figures)

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T = 25.0 degrees Celsius (3 significant figures)

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:

$Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15$

Therefore, for this problem,

$Temperature\ in \ K = 25.0 +273.15\\Temperature\ in \ K = 298.15$

Solving for n using the Ideal Gas Equation:

$n \ = \frac{PV}{RT}\\n \ = \frac{(2.00 \ atm) \ (3.00 \ L)}{(0.08206 \ \frac{L-atm}{mol-K})( 298.15 \ K)} \\n \ = 0.245 \ mol$

The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.

Learn More

1. Learn more about Boyle's Law brainly.com/question/1437490

2. Learn more about Charles' Law brainly.com/question/1421697

3. Learn more about Gay-Lussac's Law brainly.com/question/6534668

Keywords: Ideal Gas Law, Volume, Pressure

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Harlamova29_29 [7]

Answer:

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