Question

60) A student studied the kinetics of the reaction C2H4(g) C2H2(g) + H2(g) ? by monitoring the C2H4 concentration (in moles/L) as a function of time. The student made a plot of In[C2H4] against t. The plot was a straight line with the equation y =-0.61 8x-1 .427 (where y = In [C2H4] and x = t). What was the initial C2Ha concentration? a) 0.62 moles/L b) 4.17 moles/L c) 0.24 moles/L d) 0.78 moles/L e .62 moles/L

Answer 1

"Given reaction C_2 H_4 (g) rightarrow C_2 H_2 (g) + H_2 (g) �The first order rate law is ln[A] = -k t + ln[A]_0 �where [A] = concentration of 'A' at time ""t"" �k = rate constant �[A]_0 = initial concentration. �Here the order of reaction I = 1[first order] �ln[C_2 H_4] = -k t + ln[C_2 H_4]_0 �The plot ln[C_@ H_4] vs time. gave straight line with equation �y = -0.618x - 1.427 �therefore Rate constant k = 0.618 time^-1 �ln[C_2 H_4]_0 = -1.427 �[C_2 H_4]_0 = e^-1.427 �[C_2 H_4]_0 = 0.24 moles/L"