Question

InCl(s) dissolves in HCl, producing In^+​+​​. The cation, in turn, undergoes the following unbalanced reaction:

In^+​+​​(aq) → In(s) + In^{3+}​3+​​(aq).

The reaction follows first order kinetics and has a half-life of 667 s. Determine the concentration of In^+​+​​(aq) after 1.10 hrs if 2.38 g InCl(s) was dissolved in 250 mL HCl(aq). Use “E” for scientific notation.

Answer 1

Molar mass of InCl,
MM = 1*MM(In) + 1*MM(Cl)
= 1*114.8 + 1*35.45
= 150.25 g/mol


mass(InCl)= 2.38 g

number of mol of InCl,
n = mass of InCl/molar mass of InCl
=(2.38 g)/(150.25 g/mol)
= 1.584*10^-2 mol
volume , V = 250 L


Molarity,
M = number of mol / volume in L
= 1.584*10^-2/2.5*10^2
= 6.336*10^-5 M

This is initial concentration

Given:
Half life = 667 s
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(667)
= 1.039*10^-3 s-1
we have:
[In2+]o = 6.336*10^-5 M
t = 1.10 hr = 1.10*3600 s = 3960 s
k = 1.039*10^-3 s-1

use integrated rate law for 1st order reaction
ln[In2+] = ln[In2+]o - k*t
ln[In2+] = ln(6.336*10^-5) - 1.039*10^-3*3960
ln[In2+] = -9.6667 - 1.039*10^-3*3960
ln[In2+] = -13.781
[In2+] = 1.035*10^-6 M

Answer: 1.035*10^-6 M