Home
Class 12
CHEMISTRY
The rate of a reaction is expressed in d...

The rate of a reaction is expressed in different ways as follows:
`+(1)/(2)(d[C])/(dt)=-(1)/(3)(d[D])/(dt)=+(1)/(4)(d[A])/(dt)=-(d[B])/(dt)`
the reaction is

A

`4A+Bto2C+3D`

B

`B+3Dto4A+2C`

C

`A+BtoC+D`

D

`B+DtoC+D`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the reaction from the given rate expressions, we can follow these steps: ### Step 1: Understand the Rate Expressions The rate of a reaction can be expressed in terms of the change in concentration of reactants and products over time. The given expressions are: \[ +\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt} \] ### Step 2: Identify Reactants and Products From the expressions, we can identify that: - \(C\) and \(A\) are products (indicated by the positive signs). - \(D\) and \(B\) are reactants (indicated by the negative signs). ### Step 3: Write the Reaction Equation Using the stoichiometric coefficients from the rate expressions, we can construct the balanced chemical equation. The coefficients indicate the ratio in which the reactants and products are involved in the reaction: - For \(C\): 2 moles are produced. - For \(D\): 3 moles are consumed. - For \(A\): 4 moles are produced. - For \(B\): 1 mole is consumed. Thus, the reaction can be written as: \[ 3D + B \rightarrow 2C + 4A \] ### Step 4: Match with Given Options Now we can compare this equation with the provided options to find the correct one. ### Conclusion The correct reaction based on the stoichiometric coefficients is: \[ 3D + B \rightarrow 2C + 4A \] Thus, the correct answer is option 2. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

The rate of a reaction is expressed in different ways as follows: +(1)/(2)(d[C])/(dt)=-(1)/(2)(d[D])/(dt)=+(1)/(3)(d[A])/(dt)=-(d[B])/(dt) the reaction is

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) A reaction follows the given concentration-time graph. The rate for this reaction at 20 s will be

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) In the following reaction, xA rarr yB log.[-(d[A])/(dt)] = log.[(d[B])/(dt)] + 0.3 where negative isgn indicates rate of disappearance of the reactant. Thus, x:y is:

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) For reaction 3BrO^(ɵ) rarr BrO_(3)^(ɵ) + 2Br^(ɵ) , the value of rate constant at 80^(@)C in the rate law for -(d[BrO^(ɵ)])/(dt) was found to be 0.054 L mol^(-1)s^(-1) . The rate constant (k) for the reaction in terms of (d[BrO_(3)^(ɵ)])/(dt) is

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) The rate of formation of SO_(3) in the following reaction 2SO_(2) + O_(2) rarr 2SO_(3) is 100 g min^(-1) . Hence the rate of disappearance of O_(2) is

Find the degree of the following : (1+(ds)/(dt))^((3)/(2)) = 5(d^(2)s)/(dt^(2))

For the chemical reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) The correct option is: (a) 3(d[H_(2)])/(dt)=2(d[NH_(3)])/(dt) (b) -(1)/(3)(d[H_(2)])/(dt)=-(1)/(2)(d[NH_(3)])/(dt) (c) -(d[N_(2)])/(dt)=2(d[NH_(3)])/(dt) (d) -(d[N_(2)])/(dt)=(1)/(2)(d[NH_(3)])/(dt)

Mechanism of the reaction is: What is (a) (-d[A])/(dt) , (b) (d[A_(1)])/(dt)

For the reaction 2AtoB+3C , if -(d[A])/(dt)=k_(1)[A]^(2),-(d[B])/(dt)=k_(2)[A]^(2),-(d[C])/(dt)=k_(3)[A]^(2) the correct reaction between k_(1),k_(2) and k_(3) is :

The order of a reaction which has the rate expression (dc)/(dt) = k[E]^(3//2)[D]^(3//2) is