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rate of first order reaction is 0.04 moles (sec litres)inverse at 10 minutes and 0.003 moles (litres sec) at 20 minutes from start .then calculate half life for reaction
Dear student Reaction type : 1st order reactionFormula used :k =( 2.303/t2-t1) log([R1]/[R2])Given :T1 = 10 min and T2 = 20 min therefore T2-T1 = 10 min[R1] = 0.04 mol/L and [R2] = 0.03 mol/LSubstituting the values in the above equationk =( 2.303/10) log( 0.04/0.03)k = 0.2303 [log(0.04) - log (0.03)]k = 0.2303 [-1.3979 - (-1.5228)]k = 0.02876 /minFor the first order reaction half life t1/2 is given byt1/2 = 0.693/kWe know, k = 0.02876 we gett1/2 = 0.693/0.02876t1/2 = 24.096 min
k =( 2.303/t2-t1) log([R1]/[R2])
Given :T1 = 10 min and T2 = 20 min therefore T2-T1 = 10 min
[R1] = 0.04 mol/L and [R2] = 0.03 mol/L
Substituting the values in the above equation
k =( 2.303/10) log( 0.04/0.03)
k = 0.2303 [log(0.04) - log (0.03)]
k = 0.2303 [-1.3979 - (-1.5228)]
k = 0.02876 /min
For the first order reaction half life t1/2 is given by
t1/2 = 0.693/k
We know, k = 0.02876 we get
t1/2 = 0.693/0.02876
t1/2 = 24.096 min
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