Question

Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.64 × 10-2 M/s, what is the rate of formation of NOCl? 2 NO(g) + Cl2(g) → 2 NOCl(g)

Answer 1

`r_(NOCl)=9.28x10^(-2)M`

Step-by-step explanation:

Hello!

In this case, given the balanced chemical reaction:

`2 NO(g) + Cl_2(g) \rightarrow 2 NOCl(g)`

Since there is 1:2 mole ratio between chlorine and NOCl, based on the rate proportions, we can write:

`(1)/(-1)r_(Cl_2) =(1)/(2)r_(NOCl)`

It means that for the formation of NOCl, we obtain:

`r_(NOCl)=(2)/(-1)r_(Cl_2) \\\\r_(NOCl)=(2)/(-1)(-4.64x10^(-2)M)\\\\r_(NOCl)=9.28x10^(-2)M`

Notice that chlorine is disappearing, which means its rate is negate.

Best regards!