The pH of pure water at 298K and 308K are 7 and 6 respectively. Calculate the heat of formation of water from H^ + and OH^ – .

Correct answer: B. Δ H = - 84.551kcal / mol At 298K;[ H^ + ] = 10 ^ – 7 ∴ K _ w _ 1 = 10 ^ - 14 At 308K;[H^ + ] = 10 ^ – 6 ∴ K _ w _ 2 = 10 ^ - 12 Now using 2.303 log_ 10 Kw_ 2 Kw_ 1 = Δ H R [ T _ 2 - T _ 1 T _ 1 × T _ 2 ] 2.303 log _ 10 10 ^ - 12 10 ^ - 14 = Δ H 2 [(308 - 298)/(298 × 308)] ∴ Δ H = 84551.4kcal / mol = 84.551kcal / mol Thus H _ 2 O ⇌ H^ + + OH^ - ; Δ H = 84.551kcal / mol ∴ H^ + + OH^ - ⇌ H _ 2 O; Δ H - 84.551kcal / mol

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The pH of pure water at 298K and 308K are 7 and 6 respectively....

The

\text{pH}

of pure water at

\text{298}\text{K}

and

\text{308}\text{K}

are

\text{7}

and

\text{6}

respectively. Calculate the heat of formation of water from

H^{+}

and

OH^{–}

Options

A $\Delta H = 84.551kcal / mol$
B $\Delta H = - 84.551kcal / mol$
C $\Delta H = 44.981kcal / mol$
D $\Delta H = - 44.981kcal / mol$

Correct answer

B. $\Delta H = - 84.551kcal / mol$

Step-by-step solution

\text{298}\text{K;}[ H^{+} ] = {10}^{– 7}

∴ {K}_{{w}_{1}} = {10}^{- 14}

\text{308}\text{K;}\left[H^{+}\right] = {10}^{– 6}

∴ {K}_{{w}_{2}} = {10}^{- 12}

Now using

2.303{\text{log}_{10} \dfrac{{\text{Kw}_{2}}{{\text{Kw}_{1}} = \dfrac{\text{\Delta H}}{R} \left[\dfrac{{T}_{2} - {T}_{1}}{{T}_{1} \times {T}_{2}}\right]<br />2.303{\text{log}

}_{10}

\dfrac{{10}^{- 12}}{{10}^{- 14}} = \dfrac{\text{\Delta H}}{2} \left[\dfrac{308 - 298}{298 \times 308}\right]<br />

∴

\Delta

H = 84551.4kcal / mol

<br />

= 84.551kcal / mol

<br />Thus

{H}_{2}O ⇌ H^{+} + OH^{-};

\Delta

H = 84.551kcal / mol

<br />

∴ H^{+} + OH^{-} ⇌ {H}_{2}O;

\Delta

H - 84.551kcal / mol$$

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