CH3CHO decomposes into methane gas and carbon monoxide gas. This is a 2nd order reaction. Rate of decomposition at 140 degrees Celsius is .10M/s when the concentration of CH3CHO is .010. What is the rate of the reaction when the concentration of CH3CHO is .50M?
Have not been able to figure this out, but the answer is 2.5 x10^2 mol L-1 s-1
Please help and show work!
rate for second order rxn = k(A)^2
0.1 = k(0.01)^2
Solve for k.
Then substitute k into
rate = k(A)^2 with A = 0.5 and solve for rate. The answer is 250 which i 2.5E2.
To determine the rate of the reaction when the concentration of CH3CHO is 0.50 M, we can use the rate equation for a second-order reaction:
Rate = k * [CH3CHO]^2
Given that the initial rate (at 0.010 M) is 0.10 M/s, we can use it to find the rate constant (k).
0.10 M/s = k * (0.010 M)^2
k = 0.10 M/s / (0.010 M)^2
k = 1000 s^-1 M^-1
Now we can use the rate equation with the new concentration of CH3CHO (0.50 M) to calculate the rate:
Rate = (1000 s^-1 M^-1) * (0.50 M)^2
Rate = 1000 s^-1 M^-1 * 0.25 M^2
Rate = 250 M^-1 s^-1
So the rate of the reaction when the concentration of CH3CHO is 0.50 M is 250 M^-1 s^-1, which is equivalent to 2.5 x 10^2 mol L^-1 s^-1.
To solve this problem, we will use the rate law equation for a second-order reaction:
Rate = k[CH3CHO]^2
Given the rate of decomposition at 140 degrees Celsius, we can substitute the values into the equation:
0.10 M/s = k(0.010 M)^2
Solving for the rate constant (k), we get:
k = (0.10 M/s) / (0.010 M)^2
k = 100 M^-1 s^-1
Now, we can use the rate constant to find the rate of the reaction when the concentration of CH3CHO is 0.50 M:
Rate = k[CH3CHO]^2
Rate = (100 M^-1 s^-1)(0.50 M)^2
Rate = 100 M^-1 s^-1 * 0.25 M^2
Rate = 25 M^-1 s^-1
Converting to scientific notation, the rate of the reaction when the concentration of CH3CHO is 0.50 M is:
Rate = 2.5 x 10^2 mol L^-1 s^-1