The vector cross product is anti-commutative by definition. This means that for any two vectors \( \vec{P} \) and \( \vec{Q} \):
\( \vec{P} \times \vec{Q} = -(\vec{Q} \times \vec{P}) \).
The problem states that \( \vec{P} \times \vec{Q} = \vec{Q} \times \vec{P} \).
We can substitute the anti-commutative property into the given equation:
\( \vec{P} \times \vec{Q} = -(\vec{P} \times \vec{Q}) \).
Rearranging the terms, we get:
\( 2(\vec{P} \times \vec{Q}) = \vec{0} \).
This implies that the cross product of \( \vec{P} \) and \( \vec{Q} \) must be the zero vector:
\( \vec{P} \times \vec{Q} = \vec{0} \).
The magnitude of the cross product is given by \( |\vec{P} \times \vec{Q}| = |\vec{P}||\vec{Q}|\sin\theta \), where \( \theta \) is the angle between the vectors.
For the cross product to be zero, assuming \( \vec{P} \) and \( \vec{Q} \) are non-zero vectors, we must have \( \sin\theta = 0 \).
The angles for which \( \sin\theta = 0 \) in the range \( 0^\circ \le \theta<360^\circ \) are \( \theta = 0^\circ \) and \( \theta = 180^\circ \).
The problem specifies the range as \( 0^\circ<\theta<360^\circ \).
Therefore, the only possible value for \( \theta \) is \( 180^\circ \).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)