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3 years ago

What is the rate constant of a reaction if rate = 0.2 (mol/L)/s, [A] and [B] are each 3 M, m = 1, and n = 2?

Chemistry

2 answers:

Assoli18 [71]3 years ago

7 0

Answer : The rate constant of a reaction is, $7.407\times 10^{-3}M^{-2}s^{-1}$

Solution :

The general rate law expression is,

$rate=k[A]^m[B]^n$

where,

k = rate constant

[A] and [B] are the concentrations

m is the order of reactant A and n is the order of reactant B

Now put all the given values in the above rate law expression, we get the value for rate constant.

$0.2Ms^{-1}=k\times (3M)^1\times (3M)^2$

$k=7.407\times 10^{-3}M^{-2}s^{-1}$

Therefore, the rate constant of a reaction is, $7.407\times 10^{-3}M^{-2}s^{-1}$

tino4ka555 [31]3 years ago

4 0

M= 1 and n = 2
( m+ n = 1 +2 = 3)

rate = K [A] [B]^2

0.2 = K * 3 * 3 ^2
0.2 = K * 3 * 9
K = 0.2 / 27
K = 7.408 * 10 ^ -3 m^-2 s^-1

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Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

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(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

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$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

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Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

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