Question

For the reaction 2P+Q ⇌ P_2Q, the rate of formation of $P_2Q is 0.24 $mol dm^{-3}s^{-1}$. Then the rates of disappearance of P and Q respectively are

• A. -0.48$mol dm^{-3}s^{-1}$ and -0.48 $mol dm^{-3}s^{-1}$
• B. 0.24 $mol dm^{-3}s^{-1}$ and -0.48 $mol dm^{-3}s^{-1}$
• C. -0.48$mol dm^{-3}s^{-1}$ and -0.24 $mol dm^{-3}s^{-1}$
• D. -0.12$mol dm^{-3}s^{-1}$ and -0.24 $mol dm^{-3}s^{-1}$
• E. -0.24 $mol dm^{-3}s^{-1}$ and -0.12 $mol dm^{-3}s^{-1}$

Answer 1

The Correct Option is C

Rate of Reaction: 2P + Q ⇌ P₂Q 

Given: Rate of formation of P₂Q is \( +0.24 \, \text{mol dm}^{-3} \text{s}^{-1} \)

From the stoichiometry:

\[ \text{Rate} = \frac{1}{2} \left| \frac{d[P]}{dt} \right| = \left| \frac{d[Q]}{dt} \right| = \left| \frac{d[P_2Q]}{dt} \right| = 0.24 \]

So, the rate of disappearance of P:

\[ \left| \frac{d[P]}{dt} \right| = 2 \times 0.24 = \mathbf{0.48 \, mol \, dm^{-3}s^{-1}} \]

And the rate of disappearance of Q:

\[ \left| \frac{d[Q]}{dt} \right| = 0.24 \, mol \, dm^{-3}s^{-1} \]

But disappearance rates are negative:

• \( \frac{d[P]}{dt} = \mathbf{-0.48 \, mol \, dm^{-3}s^{-1}} \)
• \( \frac{d[Q]}{dt} = \mathbf{-0.24 \, mol \, dm^{-3}s^{-1}} \)

Correct Answer: -0.48 mol dm^-3s^-1 and -0.24 mol dm^-3s^-1