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shusha [124]
3 years ago
9

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the pressure is increased? There will not be a shift in

equilibrium. There will be a shift toward the products. There will be a shift toward the reactants. Not enough information is given.
Chemistry
2 answers:
Gnesinka [82]3 years ago
8 0

Answer:

There will be a shift toward the reactants.

Explanation:

According to Le Chatelier’s Principle, when a stress is applied on a system in dynamic equilibrium, the equilibrium shifts in such a way to minimize or counteract the stress. In the given reaction:

PCl₅(g) –> PCl₃(g) + Cl₂(g)

both the reactants and products are gases. But there are higher moles of gaseous particles on the product side.  As the pressure is increased the equilibrium will shift in a direction to reduce the pressure. So the equilibrium will shift towards the reactant side as that side has less moles of gaseous substance.

Tamiku [17]3 years ago
6 0
When the pressure is increased, the equilibrium will shift to the left to offset the pressure increase. Equilibrium shifting to the left side is favored because the left side has fewer moles of gas than the number of moles of gas on the right-hand side and because it exerts less pressure. Therefore, the answer is there will be a shift toward the reactants. 
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<u>Answer:</u> The conjugate acid of HCO_3^- is H_2CO_3

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CH4 + 2O2   -------- CO2 + 2H20   (Methane with O2)

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But we are also given this,

R=8.314J/mol.K, V=1450m3/h, P=150kPa gauge, Pt=150+101kPa=251kPa, T=15C= 288K

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However from the input and complete reactions stoichiometries above, we will have,

1.  Methane 86% = 0.86 x 152kmols = 130kmols, required O2 = 2 x 130.7 = 261.44kmols

2. Ethane 8% = 0.08 x 152kmols = 12.2kmols, required O2 = 3.5 x 12.2 = 42.56kmols

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But Air:O2 = 21%: 100%

inflow Air = 377.46x 100/21= 1797.5kmols, at standard pressure and temperature.

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