Answer to Question #191716 in General Chemistry for Jedidiah Barrios

Question

Hydrogen Iodide (HI) gas decomposes into gaseous hydrogen and iodine following the second-order kinetics with a rate constant of 0.031/M.s at 400ºC. 

 

2HI(g)        H2(g) + I2(g) 

 

a.What is the half-life of the reaction if the initial concentration of HI is 0.045 M? Will the half-life be different if the initial concentration is 0.025 M? Justify your answer. 

b.How much HI (in M) remains after 5.0 Minutes if the reaction started with 0.024 M HI? 

c.How many minutes will it take for the concentration of HI to decrease from 0.045 M to 0.015 M? 

Answer 1

A. Half life of second order $=\dfrac{1}{KC_0}$

Now when $C_0=0.045$ M

Half time$=$ $\dfrac{1}{0.031\times0.045}=7.1\times10^{2}$ seconds

Now when $C_0=0.025$ M

Half time$=\dfrac{1}{0.031\times0.025}=1.2\times10^3$ seconds

Here half time changes as the half time is inversely proportional to the initial conc.

B. Now for the second order reaction we know that

$\dfrac{1}{C_t}-\dfrac{1}{C_0}=Kt$

where t=5 minutes=300 seconds

Intial conc.=0.024 M

Rate constant=0.031

Now put in the equation

$\dfrac{1}{C_t}=\dfrac{1}{0.024}+0.031\times300$

$C_t=0.02M$

C. Again putting the values in the formula

$\dfrac{1}{0.015}-\dfrac{1}{0.045}=0.031\times t$

$t=143.35$ seconds or $2.39$ minutes approximately..