Question

$A\left(g\right)\to B\left(g\right)+C\left(g\right)$

-$\frac{d\left[A\right]}{dt}=k\left[A\right]$

At the beginning of this reaction, the pressure is 100 mm Hg. After 10 min, pressure increases to 120 mm Hg. The rate constant $\left(mi{n}^{-1}\right)$ is:

• A. $\frac{2.303}{10}log\frac{120}{\left(100\right)}$
• B. $\frac{2.303}{10}log\frac{100}{\left(20\right)}$
• C. $\frac{2.303}{10}log\frac{100}{\left(80\right)}$
• D. $\frac{2.303}{10}log\frac{100}{\left(120\right)}$

Answer 1

The correct option is C

$\frac{2.303}{10}log\frac{100}{\left(80\right)}$

Looking at the units of the rate constant, this is a first order reaction. We will first find out the pressure of reactant A after 10 minutes and plug that into the first order kinetics equation to get the value of the rate constant, k.

$A\left(g\right)\to B\left(g\right)+C\left(g\right)$

-$\frac{d\left[A\right]}{dt}=k\left[A\right]$

A (g) $\to$ B(g) + C(g)

100 0 0 t=0

(100-p) P P t = 10 min.

100- P+P+P= 120

(100+ P) = 120

Then k = $\frac{2.303}{10}log\frac{100}{(100-P)}$

k = $\frac{2.303}{10}log\frac{100}{\left(80\right)}$