$A + 2 B \to C$ , the rate equation for reaction is given as $R a t e = [A] [B]$ . If the concentration of A is kept constant while B is doubled. What will happen to the rate?

Doubled

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

Halved

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

The same

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

Quadrupled

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

Open in App

Solution

The correct option is C Doubled
When the concentration of A is kept constant, the rate law expression will reduce to $R a t e = k [B]$
When the concentration of B is doubled, the rate of the reaction is doubled.
$N e w R a t e = k \times 2 [B] = 2 \times o l d r a t e$

Suggest Corrections

Similar questions

A + 2 B \to C

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

A + 2 B \to C

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

^{'} B^{'}

^{'} A^{'}

2 A + 2 B \to 2 C

A + \frac{1}{2} B \to C

3 A + 2 B \to C + D,

R = K [A] [B]^{2}

A + B \to C

R = k [A] [B]^{2}

Join BYJU'S Learning Program