Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
Answer:
397 L
Explanation:
Recall the ideal gas law:
If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:
The left-hand side is simply some constant. Hence, we can write that:
Substitute in known values:
Solving for <em>V</em>₂ yields:
In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.
(Assuming 100 L has three significant figures.)
The answers are true, true, false, true, and false.
Answer:
they are big chunks of the earth that are constantly vibrating
Explanation:
Answer:
ΔG = -52.9 kJ/mol
Explanation:
Step 1: Data given
Temperature = 298 K
All species have a partial pressure of 1 atm
Δ G ° = − 69.0 kJ/mol
Step 2: The balanced equation
N2(g) + 3H2(g) ⇆ 2NH3 (g)
Step 3: Calculate Q
we will use the expression: ΔG = ΔG° + RT*ln(Q)
⇒with Q = the reaction coordinate: Q = (PNH3)²/ ((PN2)*(Ph2)³) = 666.67
Step 4: Calculate ΔG
So, ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(666.67) = -52.9 kJ/mol
(R = the gas constant = 8.314 J/mol* K OR 0.008314 kJ/mol*K)
