Question

The first-order rate constant for the decomposition of CH₃N₂CH₃, CH₃N₂CH₃(g) C₂H₆(g) + N₂(g) at 327 ^oC is k = 3.60e-04 s^-1. Suppose we start with 0.0505 mol of CH₃N₂CH₃ in a volume of 3.7 L. How many moles of CH₃N₂CH₃ will remain after 3.76e+01 min?

a) 0.0224 moles will remain.
b) 0.000684 moles will remain.
c) 1.60e-96 moles will remain.
d) 0.0498 moles will remain.
e) 0.0254 moles will remain.

Answer 1

Answer: A

Using rate of decomposition of a compound formula,

lnC(t) = lnC(0) - k*t = ln(0.0505/3.7) - 3.60e-04 * 3.76e+01 * 60 = -5.1

C(t) = 0.006

Moles remaining = 0.006 * 3.7 = 0.0224

Answer 2

Answer 3

I think that b) 0.000684 moles will remain.