Question

For the first order reaction, A¾® B.

The initial concentration of A is a₀, hence at any time t the concentration of B is x and x is given by

• A. x = a₀( 1 + e^–kt)
• B. x = a₀$\left\lbrack 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots \right\rbrack$
• C. x = a₀ $\left\lbrack - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots \right\rbrack$
• D. x = a₀ $\left\lbrack kt - \frac{(kt)^{2}}{2!} + \frac{(kt)^{3}}{3!} - \cdots \right\rbrack$

Answer 1

Answer: (D)

x = a₀ $\left\lbrack kt - \frac{(kt)^{2}}{2!} + \frac{(kt)^{3}}{3!} - \cdots \right\rbrack$

Answer 2

Rate equation is $\frac{dx}{dt}$ = k (a₀ –x)

\ ln $\frac{a_{0}}{(a_{0} - x)}$ kt

or $\frac{(a_{0} - x)}{a_{0}}$ = e^–kt

or 1 – $\frac{x}{a_{0}}$ = e^–kt

or $\frac{x}{a_{0}}$ = 1 – e^–kt

or x = a₀ (1 – e^–kt )

or x = a₀ $\left\lbrack 1–(1 - kt + \frac{(kt)^{2}}{2!} - \frac{(kt)^{3}}{3!} + \cdots) \right\rbrack$

or x = a₀ $\left\lbrack kt - \frac{(kt)^{2}}{2!} + \frac{(kt)^{3}}{3!} - \cdots \right\rbrack$