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Thermodynamics question

2022 · 29 Jun · Shift 1 · Q11
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Thermodynamics question

2022 · 29 Jun · Shift 1 · Q11

JEE MainChemistryThermodynamicsNumerical+4 / −1
17.0 g of NH3NH_3NH3​ completely vapourises at −-− 33.42 ∘^\circ∘ C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol −-− 1. The enthalpy change for the vapourisation of 85 g of NH3NH_3NH3​ under the same conditions is ‾\underline{\hspace{2cm}}​ kJ.
Numerical answer
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Correct answer: 117

  1. Given data

    • Mass of NH3NH_3NH3​ = 17.0 g17.0\,\text{g}17.0g completely vapourises
    • Enthalpy of vapourisation = 23.4 kJ mol−123.4\,\text{kJ mol}^{-1}23.4kJ mol−1
    • Required: enthalpy change for vapourisation of 85 g85\,\text{g}85g of NH3NH_3NH3​

  2. Molar mass of ammonia M(NH3)=14+3(1)=17 g mol−1M(NH_3)=14+3(1)=17\,\text{g mol}^{-1}M(NH3​)=14+3(1)=17g mol−1

  3. Find moles in 85 g of ammonia n=8517=5 moln=\frac{85}{17}=5\,\text{mol}n=1785​=5mol

  4. Use enthalpy per mole Since enthalpy of vapourisation is 23.4 kJ23.4\,\text{kJ}23.4kJ for 111 mol, for 555 mol: ΔH=5×23.4=117 kJ\Delta H = 5 \times 23.4 = 117\,\text{kJ}ΔH=5×23.4=117kJ

  5. Final answer The enthalpy change for vapourisation of 85 g85\,\text{g}85g of NH3NH_3NH3​ is 117 kJ\boxed{117\,\text{kJ}}117kJ​

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