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[ ...

299

You visited us 299 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

It will be 8 times of its initial value.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

It will be 4 times of its initial value.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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:
$r_{1} = k [S O_{2}]^{2} [O_{2}]$
$r_{2} = k [2 S O_{2}]^{2} [2 S O_{2}]$
$r_{2} = 8 r_{1}$

Suggest Corrections

Similar questions

Q. For gaseous reaction, rate

k [A] [B]

. If volume of container is reduced to

1 / 4

of initial, then the rate of reaction will be _____ times of initial.

Q. If the volume of the vessel in which the reaction

2 N O + O_{2} ⟶ {2 N O}_{2}

is occurring is diminished to 1/3rd of its initial volume. The rate of the reaction will be increased by:

Q. For gaseous reaction,

r a t e = k [A] [B]

. If volume of container is reduced to

\frac{1}{4}

of initial, then the rate of the reaction will be _________ times of initial.

A monoatomic gas $(γ = \frac{5}{3})$ is suddenly compressed to $\frac{1}{8}$ of its original volume adiabatically, then the pressure of the gas will change to.

Q. A diatomic gas

(γ = \frac{7}{5})

is compressed to

\frac{1}{32}

times of its initial volume adiabatically and reversibly. If its initial pressure is

2 P_{0}

, its new pressure will be :

Join BYJU'S Learning Program

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:
$r_{1} = k [S O_{2}]^{2} [O_{2}]$
$r_{2} = k [2 S O_{2}]^{2} [2 S O_{2}]$
$r_{2} = 8 r_{1}$