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topjm [15]
2 years ago
11

If the equilibrium concentrations are 1. 01 m a, 1. 51 m b, and 2. 05 m c. Calculate the value of the equilibrium constant of th

e reverse reaction.
Chemistry
1 answer:
pogonyaev2 years ago
3 0

0.182 is the value of the equilibrium constant of the reverse reaction.

Given reaction i.e. Forward reaction

4A + B = 3C

If we reverse this reaction, the equation becomes

3C = 4A + B

Now, the equilibrium-constant statement for the reverse reaction is the opposite of the equilibrium-constant expression for the forward reaction.

Equilibrium constant =  [A]⁴ × [B] / [C]³

The concentrations can be used to compute K(reverse) once the formula for the equilibrium constant for the reverse reaction has been established.

Conc. of A = 1.01 M

Conc. of B = 1.51M

Conc. of C = 2.05 M

Kc for reverse reaction = [1.01]⁴ × [1.51]/ [2.05]³

                                       = 0.182

Hence, 0.182 is the value of the equilibrium constant of the reverse reaction.

Learn more about equilibrium constant here brainly.com/question/12858312

#SPJ4

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What is the half-life (in seconds) of a zero-order reaction which has an initial reactant concentration of 0.884 M with a k valu
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<span>Answer: 8.15s
</span><span />

<span>Explanation:
</span><span />

<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />

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<span>=> -d[N]/dt = k [N]
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<span>2) When you integrate you get:
</span><span />

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<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />

<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />

<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
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