How would I go about solving this problem? Can someone show me an equation that can be used?
The decomposition of dimethyl ether at 504°C is a first-order reaction with a half-life of 27 min.
(CH3)2O(g) CH4(g) + H2(g) + CO(g)
(a) What will be the partial pressure of (CH3)2O(g) after 2.28 h if its initial partial pressure was 679 mm Hg?
(b) What will be the total gas pressure after 2.18 h? [Hint: (CH3)2O and its decomposition products are the only gases present in the reaction vessel.]
Can someone show me how to solve these problems (the equation I have to use)
For a you can use the first order decay constant. First, determine k from
k = 0.693/t1/2
Then substitute Po, k, and t into
ln(Po/P) = kt. Note: You must convert the t1/2 into hours (to use 2.28 hours in the problem) or leave the 27 min as is and convert the 2.28 hours into minutes.
After you know the partial pressure of the reactant and how much as decomposed, determine partial pressures of the products, then take the sum of all the gases.
To solve these problems, we would need to use the first-order reaction equation for the decomposition of dimethyl ether:
(CH3)2O(g) → CH4(g) + H2(g) + CO(g)
To solve part (a), we need to find the partial pressure of (CH3)2O(g) after 2.28 hours. We are given the initial partial pressure of (CH3)2O(g) as 679 mm Hg, and we need to determine the partial pressure after a given time.
The rate expression for a first-order reaction is given by the equation:
rate = k * [A]
In this equation, [A] represents the concentration of the reactant (CH3)2O(g) and k is the rate constant. The integrated rate law for a first-order reaction is:
ln([A]t/[A]0) = -kt
Where [A]0 is the initial concentration of the reactant, [A]t is the concentration at time t, k is the rate constant, and ln is the natural logarithm.
Since we are given the half-life of the reaction as 27 minutes, we can use the half-life equation to find the value of k:
t1/2 = 0.693/k
Substituting the given value of the half-life (27 min), we can solve for k:
0.693/k = 27 min
Solving for k gives us:
k = 0.693 / 27 min
Now we can use the rate equation and the given information to solve for the partial pressure of (CH3)2O(g) at 2.28 hours.
First, convert 2.28 hours to minutes:
2.28 hours x 60 min/hour = 136.8 min
Next, substitute the values into the integrated rate law equation:
ln([A]t/[A]0) = -kt
ln([A]t/679 mm Hg) = - (0.693 / 27 min) * 136.8 min
Simplifying further will give us the partial pressure of (CH3)2O(g) after 2.28 hours.
To solve part (b), we need to find the total gas pressure after 2.18 hours. Since (CH3)2O and its decomposition products are the only gases present in the reaction vessel, and we have already calculated the partial pressure of (CH3)2O(g), we can use the ideal gas law to find the total gas pressure.
The ideal gas law equation is:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the volume, number of moles, and temperature are constant, we can rewrite the equation as:
P1/V1 = P2/V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Since we are given the initial partial pressure of (CH3)2O(g) as 679 mm Hg, we can use this value and the total gas pressure to find the final pressure after 2.18 hours.
Hope this helps! Let me know if you have any further questions.