The decomposition of N₂O₅in solution in carbon tetrachloride proceeds via the reaction 2N₂O₅(soln) → 4NO₂(soln) + O₂(soln)
The reaction is first order and has a rate constant of 4.82 × 10^-³s^-¹at 64°C. If the reaction is initiated with 0.058 mol in a 1.00- L vessel, how many moles remain after 151 s?

Question

Quizplus · Accepted Answer

We can use the first-order integrated rate law:

ln([NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)]/[NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)]0) = -kt

First, we need to find [NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)]0:

n/V = C

0.058 mol / 1.00 L = 0.058 M

[NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)]0 = 0.058 M

Now, we can solve for [NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)] at t = 151 s:

ln([NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)]/[0.058]) = -(4.82 × 10S1U1P1-S1S1P0S1U1P1) x (151 s)

[NS1U1B12S1U1B0OS1U1B15S1U1B0S1U1B1 S1U1B0(soln)] = 0.028 M

Finally, we can calculate the amount remaining:

n = CV

n = (0.028 M)(1.00 L)

n = 0.028 mol

Therefore, the answer is C.

The Decomposition of N2O5 in Solution in Carbon Tetrachloride Proceeds

The Decomposition of N₂O₅in Solution in Carbon Tetrachloride Proceeds