For the decomposition of N2O5 at 298 k, the rate law is d[n2O5]/dt= k[N2O5] Starting with 2 5 moles - Chemistry - Chemical Kinetics

Question

For the decomposition of N2O5 at 298 k, the rate law is d[n2O5]/dt=
k[N2O5]
Starting with 2 5 moles of N2O5(g), in a 5 lt container how many
moles/ltof N2O5 will remain after 75 seconds if the rate const for
the reacn is 16 8 * 10 ^-3 s^-1

Geetha · Accepted Answer

From the rate lawR = K [N2O5],we can say that the reaction is a first order reactionK= 2
303t log aa-xt= 75 secK= 16
8 × 10-3 sec-1a= initial concentration = no of molesVolume = 2
55 = 0 5 mol/Lx= amount decomposed16
8 × 10-3= 2 30375 log 2 55a-x16
8 × 10-3 × 752
303 =  log 0 5a-xlog 0
5 - log (a-x) = 16
8 × 10-3 × 752
303log (a-x) =log 0 5 - 16
8 × 10-3 × 752 303a-x = 0
1419Thus the amount remaining after 75 sec = 0
1419 mol/L

For the decomposition of N2O5 at 298 k, the rate law is d[n2O5]/dt= k[N2O5]. Starting with 2.5 moles of N2O5(g), in a 5 lt container how many moles/ltof N2O5 will remain after 75 seconds if the rate const for the reacn is 16.8 * 10 ^-3 s^-1

For the decomposition of N2O5 at 298 k, the rate law is d[n2O5]/dt= k[N2O5].
Starting with 2.5 moles of N2O5(g), in a 5 lt container how many moles/ltof N2O5 will remain after 75 seconds if the rate const for the reacn is 16.8 * 10 ^-3 s^-1