In a reaction 2X+Y→X2Y, the reactant X will disappear at

For the reaction, 2 X + Y → X 2 Y rate = - 1 2 d X dt = - d Y dt = + d X 2 Y dt ⇒ - d X dt = - 2 d Y dt So , X will disappear twice the rate as that of disappearance of Y. Similarly, - d X dt = + d X 2 Y dt

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In a reaction $2 X + Y \to X_{2} Y,$ the reactant X will disappear at

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Twice the rate as that of appearance of $X_{2} Y$

Twice the rate as that of disappearance of Y

Half the rate as that of disappearance of Y

The Same rate as that of disappearance of Y

answer is C, D.

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Detailed Solution

For the reaction, $2 X + Y \to X_{2} Y$

$rate = - \frac{1}{2} \frac{d [X]}{dt} = - \frac{d [Y]}{dt} = + \frac{d [X_{2} Y]}{dt}$

$\Rightarrow - \frac{d [X]}{dt} = - 2 \frac{d [Y]}{dt}$

So , X will disappear twice the rate as that of disappearance of Y.

Similarly, $- \frac{d [X]}{dt} = + \frac{d [X_{2} Y]}{dt}$

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