A first order reaction has a rate constant of 2.303 10⁻³ ~s ⁻¹ . The time required for 40 ~g of this reactant to reduce to 10 ~g will be [Given that ₁₀ 2=0.3010 ]

Step-by-step solution

For first order reaction, t = 2.303 k a a - x Given : k =2.303 10⁻³ ~s ⁻¹, a =40 ~g , a - x =10 ~g On substituting the given values in Eq. (i), we get aligned t & = 2.303 2.303 10⁻³ 40 10 & =10^3 2^2=2 10^3 2 & =2 10^3 0.3010=602 ~s aligned Alternative method For first order reaction, t _ 1 / 2 = 0.693 k t _ 1 / 2 ( t _ 50 % )= 0.693 2.303 10⁻³ =301 ~s Also, t _ 75 % =2 t _ 50 % t _ 75 % =2 301=602 ~s