For the elementary reaction 2A- B - products- the rate of reaction is 6-0- 10-3 mol-lit-sec- The rate of reaction changes to 1-2- 10-2 mol-lit-sec when-

The correct option is D concentration of B is doubled 2A+B→ products Rate = K[A]^2[B] 6.0× 10^ 3= K[A]^2[B] #160; #160; #160; —–(1) ...

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Standard X

Chemistry

Factors Affecting the Rate of a Chemical Reaction

For the eleme...

Question

For the elementary reaction $2 A + B \to$ products, the rate of reaction is $6.0 \times 10^{- 3}$ mol/lit-sec. The rate of reaction changes to $1.2 \times 10^{- 2}$ mol/lit-sec when:

concentration of A and B are doubled

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concentration of B is doubled

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concentration of A is doubled

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concentration of B is tripled

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Solution

The correct option is D concentration of B is doubled
$2 A + B \to p r o d u c t s$
$R a t e = K [A]^{2} [B]$
$6.0 \times 10^{- 3} = K [A]^{2} [B]$ -----(1)
$1.2 \times 10^{- 2} = K [n_{1} A]^{2} [n_{2} B]$ -----(2)
$(2) \div (1)$
$2 = [n_{1}]^{2} [n_{2}]$
$n_{1} = 1, n_{2} = 2$
So, concentration of $B$ is doubled.

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For the elementary reaction 2A+B→ products, the rate of reaction is 6.0×10−3 mol/lit-sec. The rate of reaction changes to 1.2×10−2 mol/lit-sec when:

The correct option is D concentration of B is doubled2A+B→productsRate=K[A]2[B]6.0×10−3=K[A]2[B] -----(1)1.2×10−2=K[n1A]2[n2B] -----(2)(2) ÷(1)2=[n1]2[n2]n1=1, n2=2So, concentration of B is doubled.

For the elementary reaction $2 A + B \to$ products, the rate of reaction is $6.0 \times 10^{- 3}$ mol/lit-sec. The rate of reaction changes to $1.2 \times 10^{- 2}$ mol/lit-sec when: