Vapour Pressure of A Solution I study in class 12, and have a doubt in vapour pressure that my school teacher  and my friends are not able to clarify.
My doubt is the reason why the partial vapour pressure of each individual component of a solution is not equal to their respective standard vapour pressures($P°$)?
My reasoning is that consider you have a pure liquid A in a closed container, the vapour pressure of it shall be equal to $P°_A$. Now when you add another volatile liquid B, a solution should be formed; however, the equilibrium of the reaction
$$\ce{A_{liq} &lt;=&gt;  A_{vap}}$$ should not be affected since the concentration of reactants and products remains the same since B is different from A. Eventually, the reaction of B must also reach equilibrium, independently, and thus the partial pressure of B  must be equal to $P°_B$ .
However, this does not seem to be the case. Could someone please help in clarifying my doubt through the viewpoint of equilibrium(Mathematically if possible) and not the standard explanation that the surface of the solvent get's blocked by the particles of the solute, hence the vapour pressure reduces.
Regards,
Arun
This is where your logic fails:

since the concentration of reactants and products remains the same since B is different from A

As soon as you mix two components you change the concentration of both, you effectively dilute all components.
Otherwise your intuition is correct. The gases can often be assumed to act independently. This is true if the solution is ideal (that is, assuming interactions between all molecules in solution are roughly the same). Then each component equilibrates independently with its vapor. However, the vapor pressure of each component is in general not the same as it would be if it were pure. Instead, for an ideal solution the partial pressure of each component follows Raoult's law $p_i=p^*_i\chi_i$, where $p^*_i$ is the partial pressure of the pure component and $\chi_i$ is the mole fraction in solution.
The total pressure above the solution is then the sum of the partial pressures of the components (following Dalton's law, where $\chi_i^g$ is the mole fraction in the gas phase):
$$ \begin{align} p &amp;= \sum_i p_i = \sum_i \chi_i^g p \\ &amp;= \sum_i \chi_i p^*_i \end{align} $$