Q17 / 160 · TS EAMCET 2008 · Chemistry

Calculate $\Delta H$ in $\mathrm{kJ}$ for the following reaction $$ \mathrm{C}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) $$ Given that, $$ \begin{gathered} \mathrm{H}_2 \mathrm{O}(g)+\mathrm{C}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2(g) ; \\ \Delta H=+131 \mathrm{~kJ} \\ \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \\ \Delta H=-282 \mathrm{~kJ} \\ \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(g) ; \\ \Delta H=-242 \mathrm{~kJ} \end{gathered} $$

  • Correct answer
    A
    $-393$
  • B
    $+393$
  • C
    $+655$
  • D
    $-655$

Solution

Given, $$ \begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(g) ; \\ & \Delta H=131 \mathrm{~kJ} \\ & \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \\ & \Delta H=-282 \mathrm{~kJ} \\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {; } \\ & \Delta H=-242 \mathrm{~kJ} \\ & \mathrm{C}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta \mathrm{H}=\text { ? } \\ & \end{aligned} $$ On adding Eqs (i), (ii) and (iii), we get Eq (iv) $$ \begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2(g) ; \\ & \Delta H=+131 \mathrm{~kJ} \\ & \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(g) \\ & \Delta H=-282 \mathrm{~kJ} \\ & \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(g) \text {; } \\ & \Delta H=-242 \mathrm{~kJ} \\ & \mathrm{C}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \\ & \Delta H=(131-282-242) \mathrm{kJ} \\ & =-393 \mathrm{~kJ} \\ & \end{aligned} $$

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

Relevant Questions