Answer is: concentratio of H₃O⁺ ions is 4.2·10⁻³ M.<span>
Chemical reaction: HCOOH(aq) + H</span>₂O(l) ⇄ HCOO⁻(aq) + H₃O⁺(aq).<span>
c(HCOOH) = 0,1 M.
[</span>H₃O⁺] = [HCOO⁻] = x.<span>
[HCOOH] = 0,1 M - x.
</span>Ka = [H₃O⁺] · [HCOO⁻] / [HCOOH].
0,00018 = x² / (0,1 M - x).<span>
Solve quadratic equation: x = </span>[H₃O⁺] = 0,0042 M.
Answer:
Solution:-
The gas is in the standard temperature and pressure condition i.e. at S.T.P
Therefore,
V
i
=22.4dm
3
V
f
=?
As given that the expansion is isothermal and reversible
∴ΔU=0
Now from first law of thermodynamics,
ΔU=q+w
∵ΔU=0
∴q=–w
Given that the heat is absorbed.
∴q=1000cal
⇒w=−q=−1000cal
Now,
Work done in a reversible isothermal expansion is given by-
w=−nRTln(
V
i
V
f
)
Given:-
T=0℃=273K
n=1 mol
∴1000=−nRTln(
V
i
V
f
)
⇒1000=−1×2.303×2×273×log(
22.4
V
f
)
Explanation:
Answer:
The answer to your question is it is not at equilibrium, it will move to the products.
Explanation:
Data
Keq = 2400
Volume = 1 L
moles of NO = 0.024
moles of N₂ = 2
moles of O₂ = 2.6
Process
1.- Determine the concentration of reactants and products
[NO] = 0.024 / 1 = 0.024
[N₂] = 2/1 = 2
[O₂] = 2.6/ 1= 2.6
2.- Balanced chemical reaction
N₂ + O₂ ⇒ 2NO
3.- Write the equation for the equilibrium of this reaction
Keq = [NO]²/[N₂][O₂]
- Substitution
Keq = [0.024]² / [2][2.6]
-Simplification
Keq = 0.000576 / 5.2
-Result
Keq = 1.11 x 10⁻⁴
Conclusion
It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-
1.11 x 10⁻⁴ < 2400
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