Question

In the hypothetical reaction below, substance A is consumed at a rate of 2.0 mol/L·s. If this reaction is at dynamic equilibrium, at what rate will substance B be consumed?0.0 mol/L·s 1.0 mol/L·s 2.0 mol/L·s 4.0 mol/L·s

Answer 1

Answer:

C) 2.0

Explanation:

Answer 2

Answer : The correct answer for rate of consumption of B = $2.0 \frac{mol}{L*s}$

Dynamic equilibrium :

Dynamic equilibrium is state of equilibrium where reactants convert to product and product converts to reactant at equal and constant rate . The concentration may not be same but rate will be same .

This occurs when reaction is reversible type .

The hypothetical reaction is :

A ↔ B where ↔ represents reversible reaction

Rate of consumption of A is given as :

$R(a) = k * \frac{d[A]}{dt}$

Where . R(a) = rate of consumption of A

k = rate constant

[A] = concentration of A

t= time

Rate for consumption of B =

$R (b)= k * \frac{d[B]}{dt}$

R(b) = rate of consumption of B , [B] is concentration of B

Since the reaction is at dynamic equilibrium , so :

Rate of consumption of A = rate of consumption of B

Given : rate of consumption of A = $2.0 \frac{mol}{L*s}$

Hence rate of consumption of B = $2.0 \frac{mol}{L*s}$