A) 100
B) 0.002
C) 0.02
D) 200
Correct Answer: B
Solution :
Key Idea: (i) \[k=\frac{0.693}{{{t}_{1/2}}}\] (ii) For first order reaction r = k [cone. of reactant] Given \[{{t}_{1/2}}=693\,s\], [cone. of reactants] = 2 mol/L \[\therefore \] \[k=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{693}={{10}^{-3}}\] \[r={{10}^{-3}}\times 2\] \[=2\times {{10}^{-3}}=0.002\]
You need to login to perform this action.
You will be redirected in
3 sec
