For each of the following reactions, express the given rate of change of concentration of the reactants or Products in that reaction: (a) `H_(2)O_(2)

Monjur Alahi

For each of the following reactions, express the given rate of change of concentration of the reactants or Products in that reaction:
(a) `H_(2)O_(2) + 2H^(o+) + 3I^(ɵ) rarr I_(3)^(ɵ) + 2H_(2)O`, `(-d[I^(ɵ)])/(d t) = ?`, `(-d[H^(o+)])/(d t) ? `
(b) `16H^(o+) + 2MnO_(4)^(ɵ) + 10I^(ɵ) rarr 2Mn^(2+) + 8H_(2)O + 5I_(2)`, `(-d[MNO_(4)^(ɵ)])/(d t) = ?`
( c) `4NH_(3) + 5O_(2) rarr 4NO_(2) + 6H_(2)O` , `(-d[NH_(3)])/(d t) = ?`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

We have,
(a) `H_(2)O_(2)+2H^(+)+3I^(-)rarrI_(3)^(-)+2H_(2)O`
The equantity in this case is :
`-(d[H_(2)O_(2)])/(dt)=-(1)/(2)(d[H^(+)])/(dt)=-(1)/(3)(d-[I^(-)])/(dt)=(d[I_(3)^(-)])/(dt)=(1)/(2)(d[H_(2)O])/(dt)`
So,
`-(d[I^(-)])/(dt)=-3(d[H_(2)O_(2)])/(dt)=-(2)/(3)(d[H^(+)])/(dt)=3(d[I_(2)^(-)])/(dt)=(3)/(2)(d[H_(2)O])/(dt)`
and `-(d[H^(+)])/(dt)=-2(d[H_(2)O_(2)])/(dt)=(2)/(3)(d[I^(-)])/(dt)=2(d[I_(3)^(-)])/(dt)=(d[H_(2)O])/(dt)`
(b) `16H^(+)=2MnO_(4)^(-) +10I^(-)rarr2Mn^(2+)+8H_(2)O+5I_(2)`
The equality in this case :
`(1)/(16)(d[H^(+)])/(dt)=(1)/(2)(d[MnO_(4)^(-)])/(dt)=-(1)/(4)(d[I_(2)])/(dt)`
`(1)/(16)(d[H^(+)])/(dt)=(1)/(2)(d[MnO_(4)^(-)])/(dt)=-(1)/(10)(d[I^(-)])/(dt)`
`=(1)/(2)(d[Mn^(2+)])/(dt)=(1)/(8)(d[H_(2)O])/(dt)=(1)/(5)(d[I_(2)])/(dt)`
So, `-(d[MnO_(4)^(-)])/(dt)=-(1)/(8)(d[H^(+)])/(dt)=-(1)/(5)(d[I^(-)])/(dt)`
`=(d[Mn^(2+)])/(dt)=(1)/(4) (d[H_(2)O])/(dt)=(2)/(5) (d[I_(2)])/(dt)`
(c) `4NH_(3)+7O_(2)rarr4NO+_(2)=6H_(2)O`
The equality in this case is :
`-(1)/(4)(d[NH_(3)])/(dt)=-(1)/(5)(d[O_(2)])/(dt)=(1)/(6)(d[H_(2)O])/(dt)`
So, `-(d[Nh_(3)])/(dt)-(4)/(5)(d[O_(2)])/(dt)=(d[NO_(2)])/(dt)=(2)/(3)(d[H_(2)O])/(dt)`

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
For a reaction A+B `to` Products, the rate law is -Rate = `k[A][B]^(3//2)` Can the reaction be an elementary reaction? Explain.

In an elementary reaction, the number of atoms or ions which are colliding to react is known as molecularity. Had the reaction been of elementary nature (single step reaction) the...

2 Answers 1 views
Rate of reaction is given by the equation Rate = `k[A]^(2)[B]` What are the units for the rate and rate constant for the section?

Units of rate ( R) : Moles `L^(-1)s^(-1)` Units of `k:L^(2)mol^(-2)s^(-1)`

2 Answers 1 views
For the reaction `A + B rarr C + D`, doubling the concentration of both the reactants increases the reaction rate by `8` times and doubling the initia

Correct Answer - C c) The data suggests that reaction rate ( r) `propto [A]^(2)` and is `propto [B]`. `therefore r=k[A]6(2)[B]`

2 Answers 1 views
Consider a reaction `2A +B to` Products. When the concentration of B alone was doubled, the half life did not change. When concentration of A alone wa

Correct Answer - B b) According to available data, when the concentration of B alone was doubled while keeping that of A constant, `t_(1//2)` did not change. This happens for the...

2 Answers 1 views
For the reaction `A +2B to C`, the reaction rate is doubled if the concentration of A is doubled. The rate becomes four times when the concentration o

Correct Answer - C `A + 2B to C` Since the reaction rate becomes double when the concentration of A is made twice order w.r.t A=1 Since the reaction rate becomes...

2 Answers 1 views
The rate of reaction depends on the concentration of the reacting species, i.e., reactants in rate equations or rate law, consider the gaseous reactio

Correct Answer - A a) in the first experiment, Rate `propto [A]^(x)` `[3]^(3)=27 propto [3]^(x)` x=3

2 Answers 1 views
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers fo

Correct Answer - B `r=k[A]=0.693/69.3xx0.1=10^(-3) M, t_(1//2)=69.3` min

2 Answers 1 views
In a first order reaction, the concentration of the reactants is reduced to `25%` in one hour. The half-life period of the reactions is

Correct Answer - C `k=(2.030)/(t)log.((a)/(a-x))` `k=(2.303)/(1) log. ((100)/(5))=2xx0.693` `t_(1//2)=(0.693)/(k)=(0.693)/(2xx0.693)=(1)/(2)hr`

2 Answers 1 views
Consider the following statement : 1 The rate of reaction is always proportinal to the concentration of reactants. 2 The order of an elementary chemic

Correct Answer - C Statement (1) cannot be correct because the rate of zero order reactions does not depend on concentrartion of reactant

2 Answers 1 views
In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant

Correct Answer - D Consider a hypothetical change. `A+BhArrC+D` For the reaction. `K_(eq)=([C][D])/([A][B])` For the above reaction if concentration of reactions are doubled then the rate of forward reaction increases for...

2 Answers 1 views
X