Question

He first-order reaction, so2cl2 → so2 + cl2, has a half-life of 8.75 hours at 593 k. how long will it take for the concentration of so2cl2 to fall to 16.5% of its initial value? the first-order reaction, so2cl2 → so2 + cl2, has a half-life of 8.75 hours at 593 k. how long will it take for the concentration of so2cl2 to fall to 16.5% of its initial value? 22.7 hr 0.143 hr 2.28 hr 6.99 hr

Answer 1

0,693/k = t 1/2
0,693/k = 8,75h
0,693 = 8,75h×k
k = 0,0792 1/h
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C - 1
Ct - 16,5%× 1 = 0,165

lnC/Ct = k*t
ln1/0,165 = 0,0792×t
ln6,06 = 0,0792×t
1,802/0,0792 = t
t = 22,752 h ≈ 22,7 h

:)

Answer 2

The 22.7 hours of time will be taken for the concentration of reactant to fall to 16.5% of its initial value.

The time can be calculated by the formula,

$\bold { \dfrac {lnC_}{Ct} = k\times t}$

Where,

C - Initial concentration = 1

Ct - Concentration at time t $\bold {16.5\% \times 1 = 0.165}$

k - rate constant

$\bold {0.693/k = t ^1^/^2}\\\\\bold {0.693/k = 8.75 }\\\\\bold {0.693 = 8,75h \times k}\\\\\bold {k = 0.07921/hours}$

put the values in the formula,  

$\bold {ln\dfrac {1}{0.165} = 0.0792\times t}\\\\\bold {ln6.06 = 0.0792\times t}\\\\\bold {t = \dfrac {1.802}{0.0792}} \\\\\bold {t = 22.7}$

Therefore, the 22.7 hours of time will be taken for the concentration of reactant to fall to 16.5% of its initial value.

To know more about first order reaction,

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