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3 years ago

1. Which type of reaction is the reverse of a decomposition reaction?

Chemistry

1 answer:

Serjik [45]3 years ago

8 0

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Chemistry.

The reaction that Is opposite of the decomposition is

==> Synthesis Reaction.

==> Catalyst always Speeds up any reaction, so that it quickly reaches to it's Equilibrium point and gets stabilize.

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The equilibrium constant for the reaction 2NO(g)+Br2(g)⥫⥬==2NOBr(g) is Kc=1.3×10−2 at 1000 K. At this temperature does the equil

o-na [289]

Answer :

The equilibrium favors NO and $Br_2$ .

(1) The value of equilibrium constant for this reaction is, 76.9

(2) The value of equilibrium constant for this reaction is, 8.77

Explanation:

The given chemical equation is:

$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$

The value of equilibrium constant for the above equation is $K_c=1.3\times 10^{-2}$ .

The value of $K_c$ that means equilibrium lies to the left side. Thus, the equilibrium favors NO and $Br_2$ .

We need to calculate the equilibrium constant for the given equation of above chemical equation, which is:

(1) $2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_1}=\frac{1}{K_c}$

$K_{c_1}=\frac{1}{1.3\times 10^{-2}}=76.9$

Thus, the value of equilibrium constant for this reaction is, 76.9

(2) $NOBr(g)\rightarrow NO(g)+\frac{1}{2}Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_2}=(\frac{1}{K_c})^{1/2}$

$K_{c_2}=(\frac{1}{1.3\times 10^{-2}})^{1/2}=8.77$

Thus, the value of equilibrium constant for this reaction is, 8.77

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3 years ago

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