2. At 600 K the decomposition of gas ohase nitroethane (CH3CH2NO2) is first order with a rate constant of 1.9*10^-4 s^-1.
CH3CH2NO2 --> C2H4 + HNO2
a) Write out the differential and integrated rate laws for this reaction.
b) If a 10.0g sample of nitroethane is heated to 600K, what mass of nitroethane would be decomposed after 2.8 hours?
I assume you can do a from your text or notes. Look it up if you have doubts.
b.
ln(No/N) = kt
No = 10.0g
N = unknown
k = 1.9E-4 sec^-1 (but this must be converted to hours or the time below into seconds).
t = 2.8 hours.
Solve for N to find how much remains, then subtract from 10.0 to find the amount decomposed.
Worked! Thank you! - Why is it ln(No/N) = kt though? The formula I have in my notes for first order reactions is ln(N1/No) = -kt
BECAUSE THE MINUS SIGN WAS REMOVED IN THE EQUATION
a) To write out the differential rate law, you need to express the rate of the reaction in terms of the concentrations of the reactants. Since the reaction is first order with respect to nitroethane (CH3CH2NO2), the differential rate law can be written as follows:
Rate = k[CH3CH2NO2]
where k is the rate constant.
To write out the integrated rate law, you can integrate the differential rate law. Since the reaction is first order, the integrated rate law can be written as follows:
ln[CH3CH2NO2] = -kt + ln[CH3CH2NO2]0
where [CH3CH2NO2] is the concentration of nitroethane at a given time, [CH3CH2NO2]0 is the initial concentration of nitroethane, k is the rate constant, and t is the time.
b) To calculate the mass of nitroethane that would be decomposed after 2.8 hours, you need to use the integrated rate law and consider the molar mass of nitroethane.
First, convert the 10.0g of nitroethane to moles using its molar mass. The molar mass of nitroethane (C2H5NO2) is calculated as follows:
2(12.01 g/mol) + 5(1.008 g/mol) + 14.01 g/mol + 2(16.00 g/mol) ≈ 75.07 g/mol
So, 10.0g of nitroethane is equal to 10.0g / (75.07 g/mol) = 0.133 mol.
Next, use the integrated rate law to calculate the concentration of nitroethane at 2.8 hours. Plug in the values you know into the equation:
ln[CH3CH2NO2] = -kt + ln[CH3CH2NO2]0
Since you want to calculate the concentration after 2.8 hours, set t = 2.8 hours and solve for [CH3CH2NO2].
After finding the concentration of nitroethane at 2.8 hours, convert it back to mass using the molar mass:
Mass of nitroethane decomposed = concentration at 2.8 hours * molar mass of nitroethane
Remember to double-check your units and ensure they are consistent throughout the calculation.