Question #50ade

Question

1736 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T03:45:42

Solve the mass balance

For the first order reaction of $A\rightarrow Products$

$(dC_A)/dt = -kC_A$
$t = -1/k [ln(C_A)]|_(C_A0)^(C_A)$
$t = -1/k ln((C_A)/(C_(A0)))$

The given half time is when there is 50% of initial concentration present. Hence we have

$C_A = 0.5C_(A0)$

$3 = -1/k ln(0.5)$
$k = ln(2)/3$
$k = 0.231 hr^(-1)$

When 70 % of A is consumed the remaining concentration is

$C_A = (1-0.7)C_(A0)$
$C_A = 0.3C_(A0)$
$t = -1/0.231 ln(0.3)$
$t = 5.212$ hours.