Question

HI decomposes into its elements according to second-order kinetics. What concentration of HI remains from an initial concentration of 2.30 M after 4.5 hours? The rate constant, k, equals 1.6 x 10-3 M-1 hr-1.​​​​​​​

Answer 1

Answer:

Therefore (2.30-0.03747)= 2.2625 M of HI remains from an initial concentration of 2.30 M after 4.5 hours.

Step-by-step explanation:

second order reaction: The rate of reaction  proportional to the square of the concentration of reactant.

For second order reaction

$k=\frac{x}{ta(a-x)}$

K is rate constant = 1.6×10⁻³M⁻¹hr⁻¹

a = initial concentration of reactant = 2.30 M

a-x = concentration of reactant after t h

t = time = 4.5 h

Putting the values in the above equation,

$1.6\times 10^{-3}= \frac{x}{4.5 \times 2.30 \times (2.30-x)}$

$\Rightarrow 16.56 \times 10^{-3}\times (2.30-x)= x$

$\Rightarrow 16.56 \times 10^{-3}\times 2.30-16.56 \times 10^{-3}\times x= x$

$\Rightarrow (1+16.56 \times 10^{-3}) x=16.56 \times 10^{-3}\times 2.30$

$\Rightarrow x = 0.03747$

Therefore (2.30-0.03747)= 2.2625 M of HI remains from an initial concentration of 2.30 M after 4.5 hours.