Solveeit Logo
Login

Question

Question: Consider the reaction: \[C{{l}_{2}}\left( aq \right)+{{H}_{2}}S\left( aq \right)\to S\left( s \rig...

Consider the reaction:
Cl2(aq)+H2S(aq)→S(s)+2H+(aq)+2Cl−(aq)C{{l}_{2}}\left( aq \right)+{{H}_{2}}S\left( aq \right)\to S\left( s \right)+2{{H}^{+}}\left( aq \right)+2C{{l}^{-}}\left( aq \right)
The rate equation for this reaction is = k[Cl2][H2S]k\left[ C{{l}_{2}} \right]\left[ {{H}_{2}}S \right]
Which of these mechanisms is/are consistent with this rate equation?
A. Cl2+H2S→H++Cl−+Cl++HS−(slow)C{{l}_{2}}+{{H}_{2}}S\to {{H}^{+}}+C{{l}^{-}}+C{{l}^{+}}+H{{S}^{-}}\left( slow \right)
Cl++HS−→H++Cl−+S(fast)C{{l}^{+}}+H{{S}^{-}}\to {{H}^{+}}+C{{l}^{-}}+S\left( fast \right)
B. H2S⇔H++HS−(fast equilibrium) Cl2+HS−→H++2Cl−+S(slow) \begin{aligned} & {{H}_{2}}S\Leftrightarrow {{H}^{+}}+H{{S}^{-}}\left( fast\text{ equilibrium} \right) \\\ & C{{l}_{2}}+H{{S}^{-}}\to {{H}^{+}}+2C{{l}^{-}}+S\left( slow \right) \\\ \end{aligned}
A. B only
B. Both A and B
C. Neither A and B
D. A only

Explanation

Solution

For us to be able to solve the question, we first need to know what is the order of a reaction. To know that, we should know the rate law. It links the rate of a reaction to the concentration of the reactants and some constant parameters.

Complete step by step solution:
- The rate law can be shown as
rate=k[A]x[B]yrate=k{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}
where A and B are reactants of the equation and k is a constant parameter. The sum of x and y gives the overall order of the reaction.
-If the rate of the equation does not depend on the reactant concentration;i.e; it is constant irrespective of the reactant concentration, then the equation is zero order.
-If the rate depends on one reactant, it is a first order reaction and if it depends on two reactants or square of 1 reactant, then it is a second order reaction.
-The rate law for the above reaction is given as
k[Cl2][H2S]k\left[ C{{l}_{2}} \right]\left[ {{H}_{2}}S \right]
So the order of the reaction is 1 with respect to both the reactants and overall order is 2.
-The mechanism of any reaction depends on the rate law. To check for the consistency of the reactions, we need to form the rate law and see if it matches the given rate law.
-For option B, we see that the reaction is given as
H2S⇔H++HS−(fast equilibrium) Cl2+HS−→H++2Cl−+S(slow) \begin{aligned} & {{H}_{2}}S\Leftrightarrow {{H}^{+}}+H{{S}^{-}}\left( fast\text{ equilibrium} \right) \\\ & C{{l}_{2}}+H{{S}^{-}}\to {{H}^{+}}+2C{{l}^{-}}+S\left( slow \right) \\\ \end{aligned}
From slow step, rate law will be given as k1[Cl2][HS]−{{k}_{1}}\left[ C{{l}_{2}} \right]{{\left[ HS \right]}^{-}} and from fast reaction, it will be
K=[H+][HS−][H2S]K=\dfrac{\left[ {{H}^{+}} \right]\left[ H{{S}^{-}} \right]}{\left[ {{H}_{2}}S \right]}
-Combining both the equations, we get the rate law for the overall reaction that can be given as

& rate={{k}_{1}}\left[ C{{l}_{2}} \right]\left[ H{{S}^{-}} \right] \\\ & \Rightarrow \text{rate=}{{k}_{1}}\left[ C{{l}_{2}} \right]{{\left( K\left[ {{H}_{2}}S \right] \right)}^{1/2}} \\\ & \Rightarrow rate=k\left[ C{{l}_{2}} \right]{{\left[ {{H}_{2}}S \right]}^{1/2}} \\\ \end{aligned}$$ -For the reactions in option A, the rate law is matching that of the given rate law in the question. So the rate law is consistent with the mechanism shown by option A. **Therefore the correct option is D.** **Note:** The order can also be found by knowing the time duration in which the reaction is being completed. The duration in which half of the reaction gets completed is called half-life of the reaction and is denoted by ${{t}_{1/2}}$.