Question

The rate law
for
a reaction
below
Rate =K{A}{B}²{C}²{D}²
If conc, of A is doubled B is halved C is doubled and D is halved. Find new rate​

Answer 1

$\huge \fbox{ \underline \blue{Answer }}$

Given Rate of Reaction:

Rate =K{A}{B}²{C}²{D}²

Also said that:

Concentration of A is doubled

Concentration of B is halved.

Concentration of C is Doubled

Concentration of D is halved

To Find:

Resultant Rate of Reaction

Rate = k[A][B]²[C]²[D]²

Rate = k[2A][B/2]²[2C]²[D²/2]

$Rate = k[2A][B/2]²[2C]²[D²/2]$    Is the resultant rate of reaction

Let us calculate how much times rate of 2 increased from rate of reaction 1

R₁ =  K{A}{B}²{C}²{D}² $\longrightarrow$ Eqn 1

R₂ =  k[2A][B/2]²[2C]²[D²/2]  $\longrightarrow$ Eqn 2

Divide equation 1 by 2

$\dfrac{R₁}{R₂ }\: = \dfrac{K[A][B]²[C]²[D]²}{K[2A][B/2]²[2C]²[D/2]²}$

$\dfrac{R₁}{R₂ }\: = \dfrac{\cancel K[A][B]²[C]²[D]²}{\cancel K[2A][B/2]²[2C]²[D/2]² }$

$\dfrac{R₁}{R₂ }\: = \dfrac{[A][B]²[C]²[D]²}{2[A] 1/2 [B]²2[C]² 1/2[D]²}$

$\dfrac{R₁}{R₂ }\: = \dfrac{\cancel{[A]} \cancel{[B]²}\cancel{[C]²}\cancel{[D]²}}{2\cancel{[A]} 1/2\cancel{[B]²} 2\cancel{[C]²} 1/2\cancel{[D]²} }$

$\dfrac{R₁}{R₂} \: = \dfrac{1\times 1\times 1\times 1}{2\times \dfrac{1}{2}\times 2\times\dfrac{1}{2} }$

$R_{1} = R_{2}$  

Hence no change in reaction takes place

Hope it helps :)