Question

One quantum is absorbed per gaseous molecule of Br2 for converting into Br atoms. If light abso
has wavelength 5000A, then the bond energy of Br2 is about..... KJ/mol (1 eV/particle = 96 kJ/mo).
(A) 119
(B) 238
(C) 357
(D) 476
​

Answer 1

Answer:

The correct answer is option B.

Explanation:

$Br_2\overset{h\nu}\rightarrow 2Br$

Wavelength absorbed by the bromine gas molecule = 5000 A

$\lambda =5000 \AA= 5000\times 10^{-10} m$

Energy of the wavelength will be given by:

$E=\frac{hc}{\lambda }$

$E=\frac{6.626\times 10^{-34} J s\times 3\times 10^8 m/s}{5000\times 10^{-10} m}=3.97\times 10^{-19} Joules=3.97\times 10^{-22} kJ$

For one mole bromine gas:

$3.97\times 10^{-22} kJ\times 6.022\times 10^{23} kJ/mol=238 kJ/mol$

The correct answer is option B.

Answer 2

Br_2\overset{h\nu}\rightarrow 2Br

Wavelength absorbed by the bromine gas molecule = 5000 A

\lambda =5000 \AA= 5000\times 10^{-10} m

Energy of the wavelength will be given by:

E=\frac{hc}{\lambda }

E=\frac{6.626\times 10^{-34} J s\times 3\times 10^8 m/s}{5000\times 10^{-10} m}=3.97\times 10^{-19} Joules=3.97\times 10^{-22} kJ

For one mole bromine gas:

3.97\times 10^{-22} kJ\times 6.022\times