Chemistry MCQs for NEET — Practice Questions with Answers

To find the change in enthalpy for the reaction, we use the enthalpies of formation of the reactants and products. The reaction given is: $$2Al + Cr_2O_3 ightarrow 2Cr + Al_2O_3$$ The enthalpy change of the reaction ($ riangle H$) can be calculated using the formula: $$ riangle H = ext{Sum of enthalpies of formation of products} - ext{Sum of enthalpies of formation of reactants}$$ For the given reaction: Products: $Al_2O_3$ with enthalpy of formation = -1596 kJ Reactants: $Cr_2O_3$ with enthalpy of formation = -1134 kJ So, $$ riangle H = [-1596] - [-1134] = -1596 + 1134 = -462 ext{ kJ}$$ Therefore, the correct answer is -462 kJ.

The internal energy change ($ riangle U$) for a system depends only on the initial and final states and is independent of the path taken. Since the system returns to its initial state (A), the net change in internal energy is zero. This is a fundamental concept in thermodynamics: $$ riangle U_{ ext{net}} = riangle U_{ ext{A to B}} + riangle U_{ ext{B to A}} = 40 ext{ kJ} + (-40 ext{ kJ}) = 0$$ Hence, the net change in internal energy is zero.

To find the equilibrium constant (K) for the reaction, we use the relationship between the standard Gibbs free energy change ($ riangle G^ ext{°}$) and the equilibrium constant (K): $$ riangle G^ ext{°} = -RT ext{ln}(K)$$ Given: $$ riangle G^ ext{°} = -4.606 ext{ kcal} = -4606 ext{ cal}$$ $$T = 227^ ext{°C} = 227 + 273 = 500 ext{ K}$$ $$R = 2.0 ext{ cal K}^{-1} ext{ mol}^{-1}$$ Substituting these values into the equation, we get: $$-4606 = -2.0 imes 500 imes ext{ln}(K)$$ Solving for K: $$ ext{ln}(K) = rac{4606}{2.0 imes 500} = 4.606$$ $$K = e^{4.606} hickapprox 100$$ Therefore, the equilibrium constant (K) is approximately 100.

The given reaction is H_2(g) + I_2(g) → 2HI with ΔH = 12.40 kcal. This means that the formation of 2 moles of HI releases 12.40 kcal. Therefore, the heat of formation of 1 mole of HI is half of this value, which is 12.40 kcal / 2 = 6.20 kcal. Hence, the correct option is 6.20 kcal.

The heat of combustion of carbon monoxide (CO) to carbon dioxide (CO_2) can be determined by subtracting the heat of formation of CO from the heat of formation of CO_2: ext{ΔH}_{combustion} = -94.0 ext{ kcal} - (-26.4 ext{ kcal}) = -67.6 ext{ kcal}. Therefore, the correct option is -67.6 kcal.

The heat of conversion ( ext{ΔH}{conversion}) from rhombic sulphur to monoclinic sulphur is calculated by subtracting the heat of combustion of rhombic sulphur from the heat of combustion of monoclinic sulphur: ext{ΔH}{conversion} = -71030 ext{ cal} - (-70960 ext{ cal}) = 70 ext{ cal}. Therefore, the correct option is 70 cal.

$ W = – P \triangle V $ $ = – 1 \times 105 (1 \times 10^ {–2} – 1 \times 10 ^ {–3} ) = – 1 \times 10 ^ 5 \times 9 \times 10 ^ {–3} = – 900 J $ .