A first order gas reaction
` A_2 B_(2 (g)) to 2 A_((g) ) + 2 B_((g))`
at the temperature `400^@ C` has the rate constant `k = 2.0 xx 10^(-4)s^(-1)`. What percentage of `A_2 B_2` is decomposed on heating for 900 seconds?

Explanation

For a first order reaction,
`k=(2.303 )/(t) log "" (a)/((a-x))`
Let .a. be 100, `k = 2.0 xx 10^(-4)s^(-1) ,1= 900 s`,
Substituting the values in equation (1),
`2.0 xx 10^(-4) s^(-1) = ( 2.303 )/( 900 s) log "" (a)/((a-x))`
` log (a)/((a-x)) = ( 2.0 xx 10^(-4) s^(-1) xx 900 s )/( 2.303 ) = 0.0781 `
` log ((a-x))/( a) = - 0.781 = bar(1).9219`
` ((a-x))/( a) = ` Antilog ` = bar(1) .9219 = 0.835`
`(a-x) = 0.835 xx a = 83.5`
Amount decomposed = x=a-(a - x) = 100 - 83.5 = 16.5%
Hence, 16.5% of initial concentration has changed into products.