Question

For $S{O}_{2}C{l}_{2\left(g\right)}\to S{O}_{2\left(g\right)}+C{l}_{2\left(g\right)},$ Pressure of $S{O}_{2}C{l}_2$ changed from 5 atm to 4 atm in 10 min. Then, pressure of $S{O}_{2}C{l}_2$ at the end of 30 minutes will be (in atm):

• A. 2.56
• B. 3.56
• C. 4.56
• D. 5.56

Answer 1

Answer: (A)

2.56

Here the given reaction is:

$S{O}_{2}C{l}_{2\left(g\right)}\to S{O}_{2\left(g\right)}+C{l}_{2\left(g\right)}$

Initial pressure of $S{O}_{2}C{l}_2=5\text{atm}.$

And pressure left after $10$ minutes $=\left[a-x\right]=A=4\text{atm}\text{.}$

Hence by rate law

For first order reaction

Rate constant (K) $=\frac{2.303}{t}\log \frac{\left[A_0\right]}{\left[A\right]}$

$K=\frac{2.303}{10}\log \frac{5}{4}\Rightarrow 0.2303\left[\log 5-\log 4\right]$

$=2.2303\left[0.6980-0.6020\right]$

$K=0.0223$

Now,

$t=30min$

Concentration of $S{O}_{2}C{l}_2$ at the end of this time

$K=\frac{2.303}{t}\log \frac{\left[A_0\right]}{\left[A\right]}\Rightarrow 0.0223=\frac{2.303}{30}\log \frac{5}{\left[A\right]}$

$\frac{0.0223\times 30}{2.303}=\log 5-\log A$

$0.2907=\log 5-\log A$

$\log A=0.6989-0.2907\Rightarrow 0.40\Rightarrow \left[A\right]={10}^{0.40}\Rightarrow 2.56$