The Ka for CH3COOH at 300 and 310 K are 1-8 - 10-5 and 1-805 - 10-5- respectively- The enthalpy of deprotonation for acetic acid is 51-6 cal-

The correct option is A TrueThe Van 't Hoff equation is as shown. #160; ln (K_2/K_1)= Δ H /R(1/T_2) 1/T_1 Substitute values in the above equation. #1 ...

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Standard XII

Chemistry

pH the Power of H

The Ka for ...

Question

The $K_{a}$ for $C H_{3} C O O H$ at 300 and 310 K are $1.8 \times 10^{- 5}$ and $1.805 \times 10^{- 5}$ , respectively. The enthalpy of deprotonation for acetic acid is 51.6 cal.

True

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False

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Solution

The correct option is A True
The Van 't Hoff equation is as shown.
$l n (\frac{K_{2}}{K_{1}}) = \frac{- Δ H}{R} (\frac{1}{T_{2}}) - \frac{1}{T_{1}}$
Substitute values in the above equation.
$l n (\frac{1.805 \times 10^{- 5}}{1.8 \times 10^{- 5}}) = \frac{- Δ H}{2} (\frac{1}{310} - \frac{1}{300})$
$- Δ H = 51.6$ cal

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The Ka for CH3COOH at 300 and 310 K are 1.8×10−5 and 1.805×10−5, respectively. The enthalpy of deprotonation for acetic acid is 51.6 cal.

The correct option is A TrueThe Van 't Hoff equation is as shown.ln(K2K1)=−ΔHR(1T2)−1T1Substitute values in the above equation.ln(1.805×10−51.8×10−5)=−ΔH2(1310−1300)−ΔH=51.6 cal

The $K_{a}$ for $C H_{3} C O O H$ at 300 and 310 K are $1.8 \times 10^{- 5}$ and $1.805 \times 10^{- 5}$ , respectively. The enthalpy of deprotonation for acetic acid is 51.6 cal.