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mathrmS(g) + frac32 O_2(g) ightarrow SO_3(g) + 2xtext kcal mathrmSO2(mathrmg) + frac12mathrmO2(mathrmg) ightarrow mathrmSO3(mathrmg) + ytext kcal The heat of formation of mathrmSO_2(mathrmg) is given by:

Solution & Explanation

### Related Formula Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps: Delta Htextnet = sum Delta Htextproducts - sum Delta Htextreactants ### Core Logic The heat of formation of mathrmSO_2(g) corresponds to the target thermochemical equation: textTarget: mathrmS(g) + mathrmO2(g) ightarrow mathrmSO2(g) quad Delta H_f = ? Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so Delta H = -Q): 1. mathrmS(g) + frac32mathrmO_2(g) ightarrow mathrmSO_3(g) quad Delta H_1 = -2xtext kcal 2. mathrmSO_2(g) + frac12mathrmO_2(g) ightarrow mathrmSO_3(g) quad Delta H_2 = -ytext kcal To isolate mathrmSO_2(g) on the product side, subtract Equation (2) from Equation (1): left[mathrmS(g) + frac32mathrmO2(g) ight] - left[mathrmSO2(g) + frac12mathrmO2(g) ight] ightarrow mathrmSO3(g) - mathrmSO3(g) mathrmS(g) + mathrmO2(g) ightarrow mathrmSO2(g) Now apply the same operation to the enthalpy values: Delta H_f = Delta H1 - Delta H_2 = -2x - (-y) = y - 2xtext kcal This matches Option (2). ### Pattern Recognition To isolate your target species on the desired side of the equation, use Hess's Law to add or subtract the given elemental equations. Make sure to invert the sign of the enthalpy change if you reverse a reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

Reference Study Guides

Thermodynamics

More Thermodynamics Previous-Year Questions — Page 6

Q47 2025 Hess's Law / Enthalpy of Formation
Consider the following data: Heat of formation of CO_2(g) = -393.5mathrm~kJ~mol^-1 Heat of formation of H_2O(l) = -286.0mathrm~kJ~mol^-1 Heat of combustion of benzene = -3267.0mathrm~kJ~mol^-1 The heat of formation of benzene is ______ mathrmkJ~mol^-1 (Nearest integer).
Numerical Answer. Answer: 48 to 48

Solution

### Related Formula Enthalpy of reaction from enthalpy of formation data: Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants) ### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (C_6H_6): C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l) Given parameters: - Delta H_c = -3267.0mathrm\ kJ/mol - Delta H_f[CO_2] = -393.5mathrm\ kJ/mol - Delta H_f[H_2O] = -286.0mathrm\ kJ/mol - Delta H_f[O_2] = 0mathrm\ kJ/mol ### Step 1: Applying Hess's Law Substitute these values into the reaction expression: -3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6] -3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6] -3267 = -3219.0 - Delta H_f[C_6H_6] Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol ### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO_2 by 6 and H_2O by 3) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q19 2025 Adiabatic Process
The workdone in an adiabatic change in an ideal gas depends upon only : [cite: 1, 2]
  • A. change in its pressure
  • B. change in its specific heat
  • C. change in its volume
  • D. change in its temperature

Solution

### Related Formula Delta W = -Delta U = -n C_v Delta T ### Core Logic In an adiabatic system, no heat exchange occurs (Q=0). By the first law of thermodynamics, Delta W = -Delta U. Since internal energy U depends explicitly on temperature metrics, the total work output shifts uniquely based on temperature variation Delta T. ### Chapter Mix Class 11 Physics: Thermodynamics

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