Practice free Chemical Thermodynamics (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Which one of the following is correct ?
In terms of energy units, the correct order is: $1 ext{ cal} > 1 ext{ joule} > 1 ext{ erg}$. Specifically, $1 ext{ cal} = 4.184 ext{ joules}$ and $1 ext{ joule} = 10^7 ext{ ergs}$. Therefore, $1 ext{ cal} > 1 ext{ joule} > 1 ext{ erg}$.
When 2 moles of water is boiled at $ 100 ^ \circ C $ temperature which gets converted to vapour at same temperature . Then what will be change in entropy of system ?
For 2 moles of water vapour, Absorbed energy by system is $ \triangle H_{vap} = 2 \times 9720 = 19440 cal $ $ \triangle S _ {vap} = { \triangle H_{vap} \over T_b } $ $ = { 19440 \over (100 + 273) } $ $ = 52.12 Cal. K^ {-1} mole ^ {-1} $ $ = 52.12 4.184 $ $ = 217.6 joule K ^ {-1} . mole ^ {-1} $
Heat of vapopurisation of benzene is $ 7350 calorie K ^ {-1} mol ^ {-1} $ . Calculate the change in entropy to convert 1 mole gaseous benzene tp liquid benzene at $ 77 ^ \circ C $
What is the value of $ G ^ \circ at 25 ^ \circ C $ for the reaction having equilibrium constant 4.0 ? $ C_2 H_5 OH + CH_3 COOHCH_3 COOC_2 H_5 + H_2 O $
$ \triangle G ^ \circ = –2.303 RT logk$ $ = –2.303 \times 1.987 \times 298 log4 $ $ = –1363.7 log 4 $ $ = –1363.7 \times 0.6021$ $ = –821.1 Cal $ .
Standarrd cell potential of electrochemical is 1.20 volt for given reaction . Calculate change in free energy in KJ associated with it. $ 2 Ag ^ + + Cd \rightleftharpoons 2 Ag + Cd ^ { +2 } $
$ \triangle G ^\circ = –nFE ^\circ cell = –2 \times 96500 \times 1.20J = –231.6 KJ $
In reaction $ x \rightarrow y \triangle$ H = 4 Kcal / mol and $\triangle$ S = 10 cal /mol $k^ {-1} $ then at what temperature reaction will be spontaneously ?
$ \triangle G = \triangle H – T \triangle S $ $ O = 4000 – T \triangle 10 because ( \triangle G = 0) $ T = 400 k i.e. At 400 k, temperature reaction will be in equilibrium. But at temperature higher than 400 k the value of $ \triangle G $ will be negative. Thus at 500 k temperature reaction will be spontaneous.
If ΔH is the change in enthalpy and ΔE the change in internal energy accompanying a gaseous reaction
For a gaseous reaction, the relationship between change in enthalpy (ΔH) and change in internal energy (ΔE) is given by the equation: ΔH = ΔE + ΔngRT, where Δng is the change in the number of moles of gas. When the number of moles of the products is less than the number of moles of the reactants, Δng is negative, making ΔH less than ΔE.
The heat of reaction at constant pressure is given by
The heat of reaction at constant pressure is represented by the change in enthalpy (ΔH). It is calculated as the difference in enthalpy between the products and the reactants. Therefore, it is given by ΔH = HP - HR.
2.2016 gm of acetaldehyde produced 13.95 kcal of heat on combustion in $ O_2 $ . Calculate the heat of combustion of $ CH_3 CHO $
$ \triangle H = { 13.95 \times 44 \over 2.2016 } = 278.7 kcal $
Calculate equilibrium constant for a reaction $ 2 Br ^ {-1} + I_2 \rightleftharpoons 2 I ^ { -1} + Br_2 E ^ \circ _{cell} = -0.544 V $
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