Question

The decomposition of sulfuryl chloride into sulfur dioxide and chlorine SO2Cl2(g) → SO2(g) + Cl2(g) follows first-order kinetics. At 320◦C the rate constant is 2.2 × 10−5 sec−1 . If one started with a sample containing 0.16 moles of sulfuryl chloride per liter at 320◦C, what concentration would be left after 6.00 hours?

Answer 1

Answer: The concentration of sulfuryl chloride left is 0.0995 M

Explanation:

For the given chemical equation:

$SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant  = $2.2\times 10^{-5}s^{-1}$

t = time taken for decay process = 6.00 hours = (6 × 3600) = 21600 s    (Conversion factor:  1 hr = 3600 seconds)

$[A_o]$ = initial amount of the sample = 0.16 moles

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

$2.2\times 10^{-5}=\frac{2.303}{21600}\log\frac{0.16}{[A]}$

$[A]=0.0995moles$

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution}}$

Moles of sulfuryl chloride left = 0.0995 moles

Volume of solution = 1 L

$\text{Concentration of sulfuryl chloride left}=\frac{0.0995mol}{1L}=0.0995M$

Hence, the concentration of sulfuryl chloride left is 0.0995 M