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Ad libitum [116K]
3 years ago
8

Acetic acid (CH 3 COOH) reacts with water to form the acetate ion and the hydronium ion: CH 3 COOH(aq)+H 2 O(l) leftrightarrow C

H 3 COO^ - (aq)+H 3 O^ + (aq) At equilibriumthe concentration of CH 3 COOH is 2.0 * 10 ^ - 1 * M the concentration of CH 3 COO^ - 1.9 * 10 ^ - 3 * M and the concentration of H 3 O^ + is 1.9 * 10 ^ - 3 * M What is the value of K eq for this reaction? 1.8 * 10 ^ - 5 5.5 * 10 ^ 4 9.5 * 10 ^ - 3 1.1 * 10 ^ 2
Chemistry
1 answer:
Bezzdna [24]3 years ago
3 0

Answer: 1.8 * 10 ^ -5

Explanation:

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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
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Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

-\frac{d[B]}{dt}=k[B] - - -  -\frac{d[B]}{[B]}=k*dt

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.  

-\frac{d[P(B)]}{P(B)}=k*dt  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333

t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

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