Question

Consider the following reaction,
$2{NO}_2\left(g\right)+F_2\left(g\right)⟶2{NO}_{2}F\left(g\right)$.

The expression for the rate of reaction in terms of the rate of change of partial pressure of reactant and product is/are:

• A. rate = $-\frac{1}{2}\left[\frac{d{P}_{\left({NO}_2\right)}}{dt}\right]$
• B. rate = $\frac{1}{2}\left[\frac{d{P}_{\left({NO}_2\right)}}{dt}\right]$
• C. rate = $-\frac{1}{2}\left[\frac{d{P}_{\left({NO}_{2}F\right)}}{dt}\right]$
• D. rate = $\frac{1}{2}\left[\frac{d{P}_{\left({NO}_{2}F\right)}}{dt}\right]$

Answer 1

Answer: (A), (D)

rate = $\frac{1}{2}\left[\frac{d{P}_{\left({NO}_{2}F\right)}}{dt}\right]$
rate = $-\frac{1}{2}\left[\frac{d{P}_{\left({NO}_2\right)}}{dt}\right]$

$2{NO}_2\left(g\right)+F_2\left(g\right)⟶2{NO}_{2}F\left(g\right)$

Rate of reaction, $r=-\frac{1}{2}\left[\frac{d{P}_{\left({NO}_2\right)}}{dt}\right]$

$=-\left[\frac{d{P}_{{(F}_2)}}{dt}\right]$

$=+\frac{1}{2}\left[\frac{d{P}_{\left({NO}_{2}F\right)}}{dt}\right]$