When the chemical reaction 2NO(g)+O2(g)→2NO2(g) is carried out under certain conditions, the rate of disappearance of NO(g) is 5.0×10−5 Ms−1. What is the rate of disappearance of O2(g) under the same conditions?

A: Because two molecules of NO are consumed per molecule of O2, the rate of disappearance of O2(g) is 2.5×10−5 Ms−1.

B: Because two molecules of NO are consumed per molecule of O2, the rate of disappearance of O2(g) is 1.0×10−4 Ms−1.

C: Because the rate depends on [NO]2 and [O2], the rate of disappearance of O2(g) is 2.5×10−9 Ms−1.

D: Because the O2(g) is under the same experimental conditions as NO, it is consumed at the same rate of 5.0×10−5 Ms−1.

Option A is correct.

Because two molecules of NO are consumed per molecule of O2, the rate of disappearance of O2(g) is 2.5×10−5 Ms−1.

Explanation:

2NO(g) + O2(g) → 2NO2(g)

The rate of a chemical reaction is defined as the amount of reactant used up or the amount of products formed in the reaction.

From the chemical balance, 2 moles of NO gives 1 mole of O2, hence, it is evident that 2 molecules of NO is consumed per 1 molecule of O2.

Hence, the rate of consumption of O2 is exactly half of the rate of consumption of NO.

Rate of consumption of O2 = (1/2) × Rate of consumption of NO = (1/2) × 5.0×10−5 Ms−1

= (2.5×10−5) Ms−1

Hope this Helps!!!

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"