# How long will it take for the [HI] to drop from 1 M to 0.20 M in the zero order reaction: HI ---> 1/2 H2 + 1/2 I2 if k = 0.50 M/sec?

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### 2 Answers

The reaction given is,

HI --------->1/2H2+1/2I2

For a zero order reaction, the rate of reaction is can be expressed as,

r = k

**The rate of reaction of a zeroth order reaction is independent of the concentrations.**

k = 0.5 M/s

Therefore we can write the rate of reaction or the rate at HI reacts as follows,

r = -d[HI]/dt.

For our example,

-d[HI]/dt = 0.5 M/s

d[HI] = -0.5dt

Integrating wrt t,

[HI] = -0.5t+c

At t=0, [HI] = 1 M, then,

1 = 0+c

c = 1

Therefore the equation which gives us the [Hi] wrt time is,

[HI] = -0.5t+1

Time needed to reduce to 0.2 M can be found as,

0.2 = -0.5t +1

0.5t = 0.8

**t = 1.6 seconds**

**Therefore it will take 1.6 seconds to [HI] to drop from 1M to 0.2 M.**

**Sources:**

HI --------->1/2H2+1/2I2

For a zero order reaction, the rate of reaction is given by,

r = k

**The rate of reaction of a zeroth order reaction is independent of the concentrations.**

k = 0.5 M/s

We know,

r = -d[HI]/dt.

This is the rate of reaction, in other words the rate at which HI reacts.

Therefore,

-d[HI]/dt = 0.5 M/s

Integrating wrt t,

[HI] = -0.5t+c

At t=0, [HI] = 1 M, then,

1 = 0+c

c = 1

Therefore the equation is,

[HI] = -0.5t+1

Time needed to reduce to 0.2 M is,

0.2 = -0.5t +1

0.5t = 0.8

**t = 1.6 seconds**

**Therefore it will take 1.6 seconds to [HI] to drop from 1M to 0.2 M.**

**Sources:**