NEET Practice Questions (MCQs) with Answers & Solutions

In thermodynamics, a process is called reversible when,

Under certain conditions, the value of OG for a hypothetical reaction, $ X + Y \rightarrow Z $ is greater than zero, then –

For which of the following processes will energy be absorbed –

For the combustion of 1 mole of liquid benzene at $ 25 ^\circ C $ , the heat of reaction at constant pressure is given by, $ C_6H_6(l) + 7 O_2 (g) \rightarrow 6CO_2 (g) + 3H_2O (l); OH = –780980 cal $ . What would be the heat of reaction at constant volume?

Calculate heat of the following reaction at constant pressure, $ F_2O(g) + H_2O(g) \rightarrow O_2 (g) + 2HF(g) $ The heats of formation of $F_2O (g), H_2O(g) and HF (g) $ are 5.5 kcal, –57.8kcal and 64.2 kcal respectively.

Calculate OH at 358 K for the reaction $ Fe_2O_3(s) + 3H_2(g) \rightarrow 2Fe (s) + 3H_2O(l) $ Given that, $ OH_{298} = – 33.29 kJ mole–1 and Cp for Fe_2O_3 (s), Fe (s), H_2O (l) and H_2 (g)$ are 103.8, 25.1, 75.3 and 28.8 J/K mole.

Ka for $CH_3COOH$ at $ 25 ^\circ C $ is $ 1.754 \times 10^{–5} $ . At $ 50 ^\circ C$ , Ka is $ 1.633 \times 10^{–5} $ What will be value of $ OS^ \circ $ for the ionisation of $CH_3COOH$?

$ C_2H_6 (g) + 3.5 O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g) $ $ \triangle S_{vap} (H_2O, l)$ = $ x_1 cal K^{–1} (boiling point + T_1) $ $ \triangle Hf (H_2O, l)$ = $ x_2$ $ \triangle Hf (CO_2)$ = $ x_3 $ $ \triangle Hf (C_2H_6) $ = $ x_4 $ Hence $ \triangle H $ for the reaction is –

$ C (s) + O_2 (g) \rightarrow CO_2, (g)$ ; OH = –94.3 kcal/mol $ CO (g) + O_2(g) \rightarrow CO_2 (g)$ ; O H = – 67.4 kcal/mol $ O_2(g) \rightarrow 2O (g)$ ; OH = 117.4 kcal/mol $ CO (g) \rightarrow C (g) + O(g)$ ; OH = 230.6 kcal/mol Calculate OH for $ C (s) \rightarrow C (g) $ in kcal/mol.

The difference between OH and OE on a molar basis for the combustion of n–octane at $ 25 ^\circ C $ would be : $ 25 ^ \circ C $