Question

What is the magnitude of delta for (Ti(H2O)6)^3+ in kj/mol?​

Answer 1

Let the wavelength absorbed by the$(Ti(H_2O)+6)^{3+}$ , λ = 500 nm

We have to find, the magnitude of delta for $(Ti(H_2O)+6)^{3+}$ in $\dfrac{kJ}{mol}$.

Solution:

We know that,

The energy of the radiation is given by:

Energy, E = $\dfrac{hc}{\lambda}$

Where, h = the plank's constant = 6.626 × $10^{-34} Js$

, c = the speed of light = 3 × $10^{8} \dfrac{m}{s}$ and

λ  = the wavelength of the light wave

E = $\dfrac{6.626\times 10^{-34} Js \times 3\times 10^{8} \dfrac{m}{s}}{500\times 10^{-9}m}$

E = 3.98 × $10^{19} J$

Therefore, the energy for 1 mole of the proton can be given as:

E" = E × 6.023 × $10^{23} mol^{-1}$

Put E = 3.98 × $10^{19} J$, we get

E" = 3.98 × $10^{19} J$ × 6.023 × $10^{23} mol^{-1}$

⇒ E" = 23.97 × $10^{4} Jmol^{-1}$

⇒ E" = 239.7 $KJmol^{-1}$

Thus, the magnitude of delta for $(Ti(H_2O)+6)^{3+}$ in "239.7 $\dfrac{kJ}{mol}$".