Answer: 1.005919176541433
Explanation:
Answer:
D) 2 NOCl(g) ⇄ 2 NO(g) + Cl₂(g); Kp = 1.7 × 10⁻²
Explanation:
In order to compare the degree of decomposition of these reactions, we have to compare the equilibrium constant Kp. Kp is equal to the partial pressure of the products raised to their stoichiometric coefficients divided by the partial pressure of the reactants raised to their stoichiometric coefficients. The higher the Kp, the more products and fewer reactants at equilibrium. Among these reactions, D is the one that has the highest Kp, therefore the one experiencing the largest degree of decomposition.
Answer:
38.4 atm
Explanation:
Data obtained from the question include:
V1 (initial volume) = 3200 L
P1 (initial pressure) = 3.00 atm
V2 (final volume) = 250.0 L
P2 (final pressure) = ?
Using Boyle's law equation P1V1 = P2V2, the final pressure can be obtained as follow:
P1V1 = P2V2
3 x 3200 = P2 x 250
Divide both side by 250
P2 = 3 x 3200/250
P2 = 38.4 atm
Therefore, the pressure of the gas if ethylene is supplied by a 250.0 L tank is 38.4 atm
The effects on the concentration of SO3 gas when the
following changes occur after initial equilibrium has been established in this
system (N.C. = no change) by adding a catalyst.
2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal
So that even though there is an addition of catalyst, no
change in reactants or products has occurred because catalyst only provides a
faster pathway for the reaction to occur.
