Final answer:
The total pressure of the gas in the vessel after 40 minutes is 118.1 atm.
Step-by-step explanation:
The decomposition of dimethyl ether at 504°C is a first-order reaction with a half-life of 27.0 min, and the reaction is described by the following equation:
(CH₃)₂O(g) → CH₄(g) + H₂(g) + CO(g)
You want to know the total gas pressure in the vessel after 40.0 min.
To solve this problem, we can use the fact that the rate of the reaction is first-order with respect to the concentration of dimethyl ether, and that the initial concentration of dimethyl ether is 100% (i.e., 725 torr).
We can start by writing the rate expression for the reaction, using the fact that the rate of the reaction is first-order with respect to the concentration of dimethyl ether:
−d[CH₃]²O/dt = k[CH₃]²O
where [CH₃]²O is the concentration of dimethyl ether, and k is the rate constant for the reaction.
We are given that the half-life of the reaction is 27.0 min, which means that the concentration of dimethyl ether decreases exponentially with time. We can write this as:
[CH₃]²O = [CH₃]²O0 * e^(-kt)
where [CH³]²O0 is the initial concentration of dimethyl ether, and e is the base of the natural logarithm.
We are also given that the initial pressure of the vessel is 725 torr, which is equal to 1 atm.
We can now use the ideal gas law to relate the pressure of the gas to the concentration of the reactants and products:
PV = nRT
where P is the pressure of the gas, V is the volume of the container, n is the number of moles of gas present, and R is the gas constant.
We know that the volume of the container is constant, so we can write:
P = nRT/V
We can now use the fact that the reaction is first-order with respect to the concentration of dimethyl ether to write:
d[CH₃]²O/dt = k[CH₃]²O
We can integrate this equation over time to find the total change in the concentration of dimethyl ether:
∫[0,40] d[CH₃]²O/dt = ∫[0,40] k[CH₃]²O dt
We can evaluate this integral by using the fact that the rate constant for the reaction is 1.32 x 10^(-3) min^(-1):
∫[0,40] d[CH₃]²O/dt = ∫[0,40] k[CH₃]²O dt = 40 x 1.32 x 10^(-3) = 52.8
This means that the concentration of dimethyl ether decreases by 52.8% over the course of 40 minutes.
We can now use the ideal gas law to find the total pressure of the gas in the vessel after 40 minutes:
P = nRT/V
We know that the number of moles of gas present is equal to the initial number of moles of dimethyl ether, which is 1 mole:
n = [CH₃]²O0 = 1 mole
We can use the ideal gas constant (R = 8.3145 J/mol·K) to find the total pressure of the gas:
P = nRT/V = (1 mole) x (8.3145 J/mol·K) x (273 K) / (1 atm) = 118.1 atm
Therefore, the total pressure is 118.1 atm.
Complete question:
The decomposition of dimethyl ether at 504°C is a first order reaction with a half-life of 27.0 min.
(CH₃)₂O(g) → CH₄(g) + H₂(g) + CO(g)
A reaction vessel is charged with dimethyl ether at an initial pressure of 725 torr, and the reaction is initiated by rapidly heating the vessel to 504°C. Assuming a negligible time lag in beginning the reaction, what will the total gas pressure be in the vessel after 40.0 min? (assume ideal behavior)