After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is 1.155 10⁻³ ~s ⁻¹ ?

Step-by-step solution

Rate constant k=1.155 10⁻³ ~s ⁻¹k= 2.303 t a (a-x) a=a,(a-x)= a 2 aligned t_ 1 / 2 &= 2.303 k a a / 2 &= 2.303 1.155 10⁻³ 2 &= 2.303 1.155 10⁻³ 0.3010 aligned aligned &= 0.693 10³ 1.155 or t_ 1 / 2 &= 0.693 k = 0.693 1.155 10⁻³ &=600 ~s aligned