Answer:
the reaction will come to a halt and the other reactant will still be present.
Answer:
kp= 3.1 x 10^(-2)
Explanation:
To solve this problem we have to write down the reaction and use the ICE table for pressures:
2SO2 + O2 ⇄ 2SO3
Initial 3.4 atm 1.3 atm 0 atm
Change -2x - x + 2x
Equilibrium 3.4 - 2x 1.3 -x 0.52 atm
In order to know the x value:
2x = 0.52
x=(0.52)/2= 0.26
2SO2 + O2 ⇄ 2SO3
Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm
Equilibrium 2.88 atm 1.04 atm 0.52 atm
with the partial pressure in the equilibrium, we can obtain Kp.
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
Answer:
Explanation: the answer is A
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