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bazaltina [42]
3 years ago
14

This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and

the partial pressures of the gases
Chemistry
1 answer:
n200080 [17]3 years ago
7 0

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

<em>Where [] are concentrations at equilibrium</em>

And Q is:

Q = [HI]² / [H2] [I2]

<em>Where [] are actual concentrations of the reactants.</em>

<em />

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

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Read 2 more answers
What pressure will be produced when 2.0 moles of N2 gas is heated to 68oC in a container that holds 1.25 of gas?
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The pressure of the nitrogen gas produced is determined as 44.77 atm.

<h3>What is the pressure of the Nitrogen gas?</h3>

The pressure of the nitrogen gas is determined from ideal gas equation, as shown below;

PV = nRT

P = nRT/V

where;

  • n is number of moles = 2 moles
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P = (2 x 0.08205 x 341)/(1.25)

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Learn more about pressure here: brainly.com/question/25736513

#SPJ1

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