Decreasing the temperature in the reaction vessel keep this reaction from shifting to form more of the product.
As we know that rate of reaction is directly proportional to the concentration of the reactant.
If we increase the concentration of H2 then the rate of reaction increases. So, we keep it constant. Therefore this option is wrong.
By removing the H₂O from the reaction vessel as it almost make no change in the reaction. This can be pursuited the reaction in which product again converted into product.
By increasing the temperature we increases the rate of reaction and equilibrium shift in the forward direction.
Thus, we concluded that by decreasing the temperature in the reaction vessel keep this reaction from shifting to form more of the product.
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Answer:
V₂ = 2.91 L
Explanation:
Given data:
Initial volume = 3.50 L
Initial temperature = 90.0°C (90+273 = 363 K)
Final temperature = 30.0 °C ( 30 +273 = 303 K)
Final volume = ?
Solution:
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
V₁/T₁ = V₂/T₂
3.50 L / 363 K) = V₂ / 303 K)
V₂ = 0.0096 L/K × 303 K
V₂ = 2.91 L
