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Solved: The forward and reverse rate constants for the
Chapter 14, Problem 115P(choose chapter or problem)
The forward and reverse rate constants for the reaction \(A(g)+B(g)\ \leftrightharpoons\ C(g)\) are \(3.6 \times 10^{-3} / M \cdot s\) and \(8.7 \times 10^{-4}\ \mathrm{s}^{-1}\), respectively, at 323 K. Calculate the equilibrium pressures of all the species starting at \(P_{A}=1.6\ \mathrm{atm}\) and \(P_{B}=0.44\ \mathrm{atm}\).
Questions & Answers
QUESTION:
The forward and reverse rate constants for the reaction \(A(g)+B(g)\ \leftrightharpoons\ C(g)\) are \(3.6 \times 10^{-3} / M \cdot s\) and \(8.7 \times 10^{-4}\ \mathrm{s}^{-1}\), respectively, at 323 K. Calculate the equilibrium pressures of all the species starting at \(P_{A}=1.6\ \mathrm{atm}\) and \(P_{B}=0.44\ \mathrm{atm}\).
ANSWER:
Step 1 of 4
The given hypothetical reaction is shown in equation (1):
\(\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g})\)
The expression for rate constant for forward reaction according to rate law is:
\(\text { rate }_{f}=K_{f}[A][B]\) ----(1)
The expression for rate constant for backward reaction according to rate law is shown below:
\(\text { rate }_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}[\mathrm{C}]\) ------(2)