To determine the equilibrium constant for the overall reaction \( X \rightleftharpoons W \), we need to combine the given reactions and their respective equilibrium constants (\( K_1, K_2, \) and \( K_3 \)) in the following sequence:
Thus, the equilibrium constant for the reaction \( X \rightleftharpoons W \) is \(8.0\).
Let's evaluate the options given:
The correct answer is 8.0.
The equilibrium constant for the net reaction $X \rightleftharpoons W$ is the product of the individual constants:
\[K = K_1 \cdot K_2 \cdot K_3.\]
Substitute values:
\[K = 1 \cdot 2 \cdot 4 = 8.\]
Final Answer:
8.0
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.