at 100C the gaseous reaction A==2B+C is observed to be first order.on starting with pure A it is found that at the end of 10min,the total pressure of the system is 176mmHg,and after a long time the total pressure of the system is 270mmHg.a).determine the initial pressure of A :b).the pressure of A at the end of 10 minutes:c).half period of the reaction.
To solve this problem, we need to use the integrated rate laws for first-order reactions. These rate laws relate the concentration of a reactant to time. In this case, we will use the equation:
ln(A₀/A) = kt
where A₀ is the initial concentration of A, A is the concentration of A at time t, k is the rate constant, and t is the time.
a) To find the initial pressure of A, we can use the ideal gas law, which states that 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. Rearranging the equation, we have:
P = (n/V) * R * T
Since we're given the temperature (100°C) and pressure (176 mmHg), we can assume the volume is constant. Therefore, the initial pressure of A is 176 mmHg.
b) To find the pressure of A at the end of 10 minutes, we'll use the integrated rate law mentioned above. We can rewrite it as:
ln(A₀/A) = -kt
Rearranging the equation to solve for A, we have:
A = A₀ * e^(-kt)
Given that the total pressure of the system at the end of 10 minutes is 176 mmHg, and the initial pressure of A is also 176 mmHg, we can substitute these values into the equation:
176 = 176 * e^(-k * 10)
Simplifying, we have:
e^(-10k) = 1
Since e^0 = 1, we can deduce that -10k = 0, or k = 0. Therefore, the concentration of A does not change over time, and the pressure of A at the end of 10 minutes is still 176 mmHg.
c) The half-life of a first-order reaction is given by:
t(1/2) = ln(2)/k
Since we've determined in the previous step that k = 0, the half-life of the reaction is undefined.