Question

A reaction SO2Cl2(g)→so2(g)+cl2(g) is first order reaction with half life of 3.15 ×10^4 s at 320°C K. What percentage of SO2Cl2 would be decomposed on heating at 320°C for 90 min?​

Answer 1

Answer:

The reaction SO₂Cl₂(g)→SO₂(g)+Cl2(g) is a first-order reaction with a half-life of 3.15 ×10⁴s at 320°C. Then 11.2% of SO₂Cl₂ would be decomposed on heating at 320°C for 90 min.

Explanation:

Given that,

half-life,$t_{1/2}$ =  3.15 ×10⁴s

We know that for first-order reactions,

                 $t_{1/2} =\frac{0.693}{k}$              

                                   where k=rate constant

⇒                  $k=\frac{0.693}{t_{1/2} } =\frac{0.693}{3.15\times 10^{4} } =2.2\times10^{-5} s^{-1}$

time given ,t = 90 min = 90×60 s= 5400 s

Integrated Rate Law for a first-order reaction can be written as:

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

here [A₀] denotes the concentration of SO₂Cl₂ at t=0. [A] denotes the concentration of at the time t=t.

assume that [A₀] = 100

then,

            $log\frac{[Ao]}{[A]} =\frac{kt}{2.303}=\frac{2.2\times 10^{-5} \times5400 }{2.303} = 0.5158$

                  $\frac{100}{[A]} =antilog(0.0516)=1.126$

                  $[A]=\frac{100}{1.126} =88.8$

after 90 min,88.8 concentration of SO₂Cl₂ in the total concentration of 100 is left.

That means,

percentage decomposition of SO₂Cl₂ = 100-88.8 = 11.2%