Answer to Question #106478 in General Chemistry for Beverlie

Question

1. The reaction 2A → A2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 4.50 M, what was the concentration of A (in M) after 180.0 min? answer in M.
2. The reaction 2A → A2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 2.50 M, what was the concentration of A (in M) after 180.0 min? answer in M.
3. The reaction 2A → A2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 2.75 M, what was the concentration of A (in M) after 180.0 min? answer in M.

Answer 1

$-dA/dt=kA^2 \implies _{A_o}^{A_t}\int dA/A^2=_0^t\int(-k)dt$

Solving this, we get;

$1/A_t-1/A_0=kt$

This is the second order rate equation.

1.Given : $k=0.0265 M^{–1}min^{–1}; A_0= 4.50 M$

Thus, $1/A_t=2/9+180*0.0265=4.999$

$\implies A_t=0.2M.$

2. Given : $k=0.0265 M^{–1}min^{–1}; A_0= 2.50 M$

Thus, $1/A_t=2/5+180*0.0265=5.17$

$\implies A_t=0.1934M.$

3.Given : $k=0.0265 M^{–1}min^{–1}; A_0= 2.75 M$

Thus, $1/A_t=2/11+180*0.0265=4.952$

$\implies A_t=0.2019M.$