प्रथम कोटि के अभिक्रिया के लिए
`t=(2.303)/Klog""a/((a-x))`
यदि `a=100," "x=20," "t=10`
`therefore" "10=(2.303)/tlog_(10)""a/((a-x))" "...(1)`
या `K=(2.303)/(10)log""(100)/(80)=(2.303)/(10)xx0.0969=0.0969=0.022" मिनट"^(-1)`
यदि `a=100," "x=75," "t=t`
`therefore" "t=(2.303)/Klog""(100)/(100-75)" "...(2)`
`t=(2.303)/(0.022)log4=(2.303)/(0.022)xx0.6020`
`t=63.018` मिनट
