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kupik [55]
3 years ago
15

Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil

ibrium with 9.56x102 MNH3(g) and 7.62x102 M H2S(g), what is the value of the equilbrium constant at 302 K? K =
Chemistry
1 answer:
quester [9]3 years ago
7 0

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

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Explanation:

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3 years ago
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Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>

For the zero-order reactions, concentration-time equation can be written as follows:

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<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>

Data:

K = 4.2 ×10−3atm/s,  

[A]o=[HBr]o= 2 atm,  

[A]=[HBr]=0 atm (all HBr(g) is gone)

<em>We clear the incognita :</em>

[A] = - Kt + [Ao]............. Kt =  [Ao] - [A]

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<em>We replace the numerical values:</em>

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

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3 years ago
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