Consider the reversible chemical reaction: A ? B. Suppose that the forward react

Question

Consider the reversible chemical reaction: A ? B.
Suppose that the forward reaction (from A to B) has a rate constant k and the
reverse reaction (from B to A) has a rate constant q. Further, suppose that
initially (at time t = 0), there are m molecules of A and zero molecules of B
present inside the reactor.
1) Formulate a model for the rate at which the number of B molecules in
the reactor change with time.
2) Solve the model to obtain an expression for the number of B-molecules in
the reactor as a function of time.
3) What will be the equilibrium number of B-molecules in the reactor?

Explanation / Answer

Let after time t, [A] = m-x and [B] = x

1) Then, rate of formation of B, d[B]/dt = dx/dt = k[A] - q[B]

= k(m-x) - qx

so dB/dt   = km - x(k+q)

2) Solving above: dx/dt = km - (k+q)x from t=0 to t=t and x=0 to x=x

=> ln[{km - (k+q)x}/km] = -(k+q)t

=> x = km[1-e-(k+q)t]/(k+q) ANS.

3) At equilibrium,

k[A] = q[B]

or, k(m-x) = qx

or, x = km/(k+q)

So, [B] at eqbm = km/(k+q) ANS

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