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

Step-by-step 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=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant =
`2.2* 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* 10^(-5)=(2.303)/(21600)\log(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}=(0.0995mol)/(1L)=0.0995M`

Hence, the concentration of sulfuryl chloride left is 0.0995 M