Q8 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Chemistry

If enthalpy of combustion of carbon to $\mathrm{CO}_2(\mathrm{~g})$ is $-394.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the enthalpy change for the formation of $17.6 \mathrm{~g}$ of $\mathrm{CO}_2$ from carbon and dioxygen at the same temperature in $\mathrm{kJ}$ is

  • Correct answer
    A
    -157.6
  • B
    315.2
  • C
    157.6
  • D
    -315.2

Solution

Reaction for combusion is $$ \mathrm{C}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2, \Delta \mathrm{H}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1} $$ $\because$ Molar mass of $\mathrm{CO}_2=12+(2 \times 10)$ and $$ =44 \mathrm{~g} $$ $\because 44 \mathrm{~g}$ of formation of $\mathrm{CO}_2$ gives heat $=-394.0 \mathrm{~kJ}$ $\therefore 17.6 \mathrm{~g}$ of formation of $$ \mathrm{CO}_2 \text {, gives heat }=\frac{-394 \times 17.6}{44}=-157.6 \mathrm{~kJ} \text {. } $$

Hence, the correct answer is option: 'A'.

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