Question

Given that:
$A(s) \to A(l);\Delta H = x$
$A(l) \to A(g);\Delta H = y$
The heat of sublimation of A will be
A. $x - y$
B. $x + y$
C. x or y
D. $- (x + y)$

Answer 1

When the matter goes from one state to the other it is observed that the temperature of the process remains the same even while there is a steady supply of heat given to the matter. It has been found out that such heat goes into the breaking of the bonds that will lead to the change in state, which is essential for the change of matter from one phase to another.

Complete step by step answer:
If we add the two reactions given in the question above we get
$A(s) \to A(l);\Delta H = x$
$A(l) \to A(g);\Delta H = y$
Upon addition,
$A(s) \to A(g);\,\Delta H = x + y$
This result is achieved as the thermodynamic equations can be added arithmetically and thus their enthalpies can also be processed as such. The compounds in the liquid state get canceled and we get the reaction that only contains the solid and the liquid states.

Thus the correct answer is (B)

Note: All processes that occur in nature tend to proceed in a single direction spontaneously. Here spontaneity means the potential to proceed without the assistance of any of the external agents.
S is called the entropy of the reaction and is the measure of the disorder. The entropy of the system is usually for the system that is isolated. the system has the default tendency to gain more randomness. The disorder is directly linked to the entropy of the system. Both the combined action of the entropy and the enthalpy is needed for the spontaneity of the process.
For a spontaneous process, the change in entropy should be greater than 0 in other words $\Delta S > 0$
In the process of reduction, the change in enthalpy of the system is less than 0 so it is endothermic.