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ale4655 [162]
2 years ago
10

A certain reaction at equilibrium has more moles of gaseous products than of gaseous reactants.(b) Write a statement about the r

elative sizes of Kc and Kp for any gaseous equilibrium.
Chemistry
1 answer:
mojhsa [17]2 years ago
6 0

To be equal as K_p , K_c needs to be divided by (RT)^{\Delta n}

<h3>Briefly explained</h3>

When the amount of gaseous reactant is greater than the amount of gaseous product, where n_{products} < n_{reactants},

then = K_p < K_c

It's because \Delta n is negative, which places (RT) in the denominator. This is how the equation will now appear.

K_p = K_c(RT)^{-\Delta n} = \frac{ K_c}{(RT)^{\Delta n}}

Here you can observe the value of K_c  is greater than K_p. To be equal as K_p , K_c needs to be divided by (RT)^{\Delta n}

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.

Learn more about equilibrium

brainly.com/question/11336012

#SPJ4

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<h3>Answer:</h3>

134 atm

<h3>Explanation:</h3>
  • Based on the pressure law, the pressure of a gas varies directly proportionally to the absolute temperature at a constant volume.
  • Therefore; we are going to use the equation;

\frac{P1}{T1}=\frac{P2}{T2}

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Initial temperature, T1 (48°C) = 321 K

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