Given - H0ioninazHCN - 45-2 kJ mol -1 and - H0ioninazCH 3COOH - 2-1 kJ mol -1- Which one of the following is true-

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$

Calculate the enthalpy change for the following reaction
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
Given enthalpies of formation of $C H_{4}, C O_{2} & H_{2} O$ are $- 74.8 K J m o l^{- 1}, - 393.5 K J m o l^{- 1}$ and $- 286.2 K J m o l^{- 1}$ respectively.

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GivenΔH0ioninaz(HCN) = 45.2 kJ mol−1 and ΔH0ioninaz(CH3COOH) = 2.1 kJ mol−1. Which one of the following is true?

The correct option is B pka(HCN)>pka(CH3COOH)HCN is a weaker acid than CH3COOH from ΔH ionization valuepkaα1strengthofacid Since HCN is weaker acid∴ pka (HCN) > pka (CH3COOH)$

Given $Δ H_{i o n i n a z}^{0}$ (HCN) = 45.2 kJ mol $^{- 1}$ and $Δ H_{i o n i n a z}^{0}$ (CH $_{3}$ COOH) = 2.1 kJ mol $^{- 1}$ . Which one of the following is true?

The correct option is B $p^{k_{a}} (H C N) > p^{k_{a}} (C H_{3} C O O H)$
HCN is a weaker acid than $C H_{3} C O O H$ from $Δ H$ ionization value
$p^{k a} α \frac{1}{s t r e n g t h o f a c i d}$
Since HCN is weaker acid
$∴$ pka (HCN) > pka (CH $_{3}$ COOH)$