For a gas reacting on a solid surface with rate r = (kKP)/(1+KP) and K = 0.50 atm^-1, the apparent order with respect to

Question

For a gas reacting on a solid surface with rate r = (kKP)/(1+KP) and K = 0.50 atm^-1, the apparent order with respect to the gas is n = 1/(1+KP). Select all correct statements.

Accepted Answer

At P = 0.020 atm the order is about 0.99.

Accepted Answer

At P = 18 atm the order is 0.10.

Accepted Answer

At P = 18 atm the rate is 90% of the limiting rate k.

Answer

At P = 2.0 atm the order is 1/3.

For a gas reacting on a solid surface with rate $r = \frac{k K P}{1 + K P}$ and $K = 0.50 at m^{- 1}$ , the apparent order with respect to the gas is $n = 1/ (1 + K P)$ . Select all correct statements.

Answer: B, C, D

More questions on Catalysis: homogeneous, heterogeneous, enzyme catalysis

For a gas reacting on a solid surface with rate r=1+KPkKP​ and K=0.50 atm−1, the apparent order with respect to the gas is n=1/(1+KP). Select all correct statements.

Answer: B, C, D

More questions on Catalysis: homogeneous, heterogeneous, enzyme catalysis

For a gas reacting on a solid surface with rate $r = \frac{k K P}{1 + K P}$ and $K = 0.50 at m^{- 1}$ , the apparent order with respect to the gas is $n = 1/ (1 + K P)$ . Select all correct statements.