How will the rate of reaction 2SO2g - O2g - 2SO3g change if the volume of the reaction vessel is halved -

The correct option is C It will be 8 times of its initial value. V = 1/2, then concentration = #160;2 times.Assuming the law rate law as: r_1 = k[ ...

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Standard IX

Chemistry

Order of Reaction

How will the ...

Question

How will the rate of reaction $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ change if the volume of the reaction vessel is halved :

It will be 1/6th of its initial value.

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It will be 1/4th of its initial value.

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It will be 8 times of its initial value.

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It will be 4 times of its initial value.

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Solution

The correct option is C It will be 8 times of its initial value.
$V = \frac{1}{2},$ then concentration = 2 times.
Assuming the law rate law as:
$r_{1} = k [S O_{2}]^{2} [O_{2}]$
$r_{2} = k [2 S O_{2}]^{2} [2 S O_{2}]$
$r_{2} = 8 r_{1}$

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How will the rate of reaction 2SO2(g)+O2(g)→2SO3(g) change if the volume of the reaction vessel is halved :

The correct option is C It will be 8 times of its initial value.V=12, then concentration = 2 times.Assuming the law rate law as:r1=k[SO2]2[O2]r2=k[2SO2]2[2SO2]r2=8r1

How will the rate of reaction $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ change if the volume of the reaction vessel is halved :

The correct option is C It will be 8 times of its initial value.
$V = \frac{1}{2},$ then concentration = 2 times.
Assuming the law rate law as:
$r_{1} = k [S O_{2}]^{2} [O_{2}]$
$r_{2} = k [2 S O_{2}]^{2} [2 S O_{2}]$
$r_{2} = 8 r_{1}$