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Katarina [22]
3 years ago
14

The equilibrium reaction below has the Kc = 0.254 at 25°C. If the temperature of the system at equilibrium is decreased to 0°C,

how and for what reason will the equilibrium shift? Also show and explain how and why the Kc value will change.
Chemistry
1 answer:
tia_tia [17]3 years ago
7 0

Answer:

The equilibrium shifts towards reagents

Explanation:

Is known Kc=\frac{[products]}{[reagents]}  rised to the power of their number of moles in the balanced reaction. When you have a system at equilibrium with Kc < 1, it means [products] < [reagents] and the system needs energy to react, so if you decrease tempeture the equilibrium shifts towards reagents and less products will be created.

This efect can be discribed with Van´t Hoff equation: ln(\frac{K1}{K2}) =-\frac{dH}{R} (\frac{1}{T1} -\frac{1}{T2} ) where we can see that if we decrease temperature (this is T2<T1) in consecuense K2<K1 and reaction doesn´t happen.

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C6H14O3F

Explanation:

The first step is to divide each compound by its molecular weight

Carbon

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Hydrogen

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Oxygen

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Flourine

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0.542/0.542

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How much CaCO3 would have to be decomposed to produce 247 g of CaO
vovangra [49]

441 g CaCO₃ would have to be decomposed to produce 247 g of CaO

<h3>Further explanation</h3>

Reaction

Decomposition of CaCO₃

CaCO₃ ⇒ CaO + CO₂

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mol of CaO(MW=56 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{247}{56}\\\\mol=4.41

From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :

\tt \dfrac{1}{1}\times 4.41=4.41

mass CaCO₃(MW=100 g/mol) :

\tt mass=mol\times MW\\\\mass=4.41\times 100\\\\mass=441~g

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3 years ago
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