Thermodynamics And Thermochemistry Result Question 13

Answer:

Correct Answer: 13. ( b, c, d )

Solution:

13. $\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2, \quad \Delta H>0$

It is reverse of combustion of $\mathrm{H}_2(\mathrm{~g})$, hence endothermic.

$\mathrm{C}_2 \mathrm{H}_6 \longrightarrow \mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2 ; \quad \Delta H>0$

Here, more stable (saturated) hydrocarbon is being transformed to less stable (unsaturated) hydrocarbon, hence endothermic.

$\mathrm{C}{(\mathrm{gr})} \longrightarrow \mathrm{C}{(\mathrm{d})}, \quad \Delta H>0$