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The reaction SO_(2)Cl_(2) to SO_(2)+Cl_(2) is a first order reaction with half-life 3.15xx10^(4)s at 320^(@)C. What percentage of SO_(2)Cl_(2) would be decomposed on heating at 320^(@)C for 90 minutes? |
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Answer» Solution :From half-life data, we can OBTAIN the value of k `t_(1//2)=(0.693)/(k) or k= (0.693)/(3.15xx10^(4)s)` Applying first order equation and substituting the values, we GET `k=(2.303)/(t)"log"([A]_(0))/([A])` or `(0.693)/(3.15xx10^(4)s)=(2.303)/(90xx60s)"log"(100)/([A])` or `"log"(100)/([A])=(0.693 XX 90 xx 60)/(2.303xx3.15xx10^(4))=(0.301xx5400)/(3.15xx10^(4))=(16.254)/(315)=0.0516` or `(100)/([A])="Antilog "0.0516=1.126` or `[A]=(100)/(1.126=88.8%` PERCENTAGE of `SO_(2)Cl_(2)` decomposed `=100-88.8=11.2%`. |
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