Chemistry MCQs for NEET — Practice Questions with Answers

First, calculate the total energy that can be provided by the 11.2 kg of butane. The molar mass of butane ( ext{C_4H_{10}}) is ext{58 g/mol}, so the number of moles in 11.2 kg is ext{11200 g / 58 g/mol = 193.1 mol}. The energy released per mole of butane is -2658 kJ, so the total energy available is ext{193.1 mol * 2658 kJ/mol = 513,679.8 kJ}. Given that a normal family needs 20000 kJ of energy per day, the number of days the cylinder will last is ext{513,679.8 kJ / 20000 kJ/day ≈ 25.68 days}. Since we are looking for an integer value, the closest option is 26 days.

The reaction given is the transition of carbon from diamond to graphite with a negative enthalpy change (ΔH = -453.5 cal). A negative ΔH indicates that the process releases heat, meaning the product (graphite) is more stable than the reactant (diamond). Hence, graphite is more stable than diamond.

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, $1 ext{ cal} > 1 ext{ joule} > 1 ext{ erg}$.

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} $

$ \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 $ .

$ \triangle G ^\circ = –nFE ^\circ cell = –2 \times 96500 \times 1.20J = –231.6 KJ $

$ \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.

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 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.