The rate of formation of B is:
\[ \frac{d[\text{B}]}{dt} = k_1[\text{A}] - k_2[\text{B}]. \]
For the rate of formation of B to be zero:
\[ \frac{d[\text{B}]}{dt} = 0. \]
Substitute:
\[ k_1[\text{A}] - k_2[\text{B}] = 0. \]
Rearrange to find [B]:
\[ k_1[\text{A}] = k_2[\text{B}] \implies [\text{B}] = \frac{k_1}{k_2}[\text{A}]. \]
Thus, the concentration of B is:
\[ [\text{B}] = \left(\frac{k_1}{k_2}\right)[\text{A}]. \]
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.