Question

In Breckenridge, Colorado, the typical atmospheric pressure is 50 torr. What is the boiling point of water (ΔHvap = 40.7 kJ/mol) in Breckenridge?


Give the expected hybridization of the central atom for the following molecules or ions.

(a) NO₂⁻
(b) SeCl₂
(c) N₃⁻
(d) NO₃⁻
(e) CH₄
(f) OCN⁻ (carbon is the central atom)

Answer 1

Solution:- First problem is based on Clausius-Clapeyron equation.

ln(P₂/P₁) = delta (H_vap/R) [1/T₁ - 1/T₂]

we know that normal boiling point of water is 100 degree C that is 373K at 760 torr. What would be the boiling point at 50 torr.

delta H_vap = 40.7 kJ/mol = 40700 J/mol, P₁ 760 torr, T₁ = 373K, P₂ = 50 torr, T₂ = ?, R = 8.314 J/(mol*K)

ln(50/760) = (40700/8.314) [1/373 - 1/T₂]

-2.72 = 4895.36 [1/373 - 1/T₂]

divide both sides by 4895.36

-5.56 x 10^-4 = [1/373 - 1/T₂]

-5.56 x 10^-4 = 0.00268 - 1/T₂

-0.000556 - 0.00268 = - 1/T₂

-0.003236 = - 1/T₂

negative sign is canceled out as it is on both sides..

0.003236 = 1/T₂

T₂ = 1/0.003236 = 309

So, the boiling point of water at 50 torr would be 309K.

(2)

two etoms ane bonded to N(central atom) and It har ono Inan pair so The 의am t53 and nance tha hybndraton ts sp2 are bonded to se andit sum is4 and hence the only twostoms ล"e bonded to the ฎentral aton and it has no Icen pair s0 the sum is 2 end hence the hybridization is sp. Hyorldization for each temeinal N atom is sp2. ol Three stoms are tonded to central N etom andithas no Iaen peir so the sum is 3 and hence Its hybridizetion is 302. : O Only towo atoms are bonded to C and it has no loan pair so the sum it 2 and hence ithybriduraton is sp