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Thermodynamics question

2021 · 22 Jul · Shift 2 · Q20
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Thermodynamics question

2021 · 22 Jul · Shift 2 · Q20

JEE MainChemistryThermodynamicsNumerical+4 / −1
If the standard molar enthalpy change for combustion of graphite powder is −-− 2.48 ×\times× 102 kJ mol −-− 1, the amount of heat generated on combustion of 1 g of graphite powder is ‾\underline{\hspace{2cm}}​ kJ. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 21

  1. Write the given data
  • Standard molar enthalpy of combustion of graphite: ΔHc∘=−2.48×102 kJ mol−1=−248 kJ mol−1\Delta H_c^\circ = -2.48 \times 10^2\ \text{kJ mol}^{-1} = -248\ \text{kJ mol}^{-1}ΔHc∘​=−2.48×102 kJ mol−1=−248 kJ mol−1
  • Mass of graphite burnt: m=1 gm = 1\ \text{g}m=1 g
  • Molar mass of graphite (carbon): M=12 g mol−1M = 12\ \text{g mol}^{-1}M=12 g mol−1
  1. Find moles of graphite in 1 g

n=mM=112 moln = \frac{m}{M} = \frac{1}{12}\ \text{mol}n=Mm​=121​ mol

  1. Calculate heat generated

For 1 mole of graphite, heat released is 248 kJ248\ \text{kJ}248 kJ. So for 112\dfrac{1}{12}121​ mole,

q=248×112=20.67 kJq = 248 \times \frac{1}{12} = 20.67\ \text{kJ}q=248×121​=20.67 kJ

Since the question asks for the amount of heat generated, we report the magnitude:

∣q∣=20.67 kJ|q| = 20.67\ \text{kJ}∣q∣=20.67 kJ

  1. Nearest integer

20.67≈2120.67 \approx 2120.67≈21

Therefore, the required answer is:

21\boxed{21}21​

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