Question

Cyclopropane, C3H6, is converted to its isomer propylene, CH2=CHCH3, when heated. The rate law is first order in cyclopropane, and the rate constant is 6.0 × 10−4/s at 500°C. If the initial concentration of cy- clopropane is 0.0226 mol/L, what is the concentration after 525 s?

Answer 1

Answer: 0.011 M

Explanation:

$C_3H_6\rightarrrow CH_2=CH-CH_3$

$Rate=k[C_3H_6]^1$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $6.0\times 10^{-4}\text{s}^{-1}$

t = time taken   = 525 sec

a = initial amount of the reactant  = 0.0226 mol/L

a - x = amount left = ?

Now put all the given values in above equation, we get

$525=\frac{2.303}{6.0\times 10^{-4}}\log\frac{0.0226}{(a-x)}$

$(a-x)=0.011mol/L$

Thus the concentration after 525 s is 0.011 mol/L.