At 518°c, the rate of decomposition of a sample of gaseousacetaldehyde, initially at a pressure of 363 torr, was 1.07 torr s−1when 5.0 percent had reacted and 0.76 torr s−1when 20.0 per cent had reacted. determinethe order of the reaction. web. bumc. bu. edu/otlt/mph-modules/bs/bs704_proba bility/bs704_probability4.html.

2nd order reaction

Explanation:

Let us assume the reaction to be:

                         R → P

Where R is the reactant and P is the product.

So here, say initially we have "a" amount of reactant.

                        R → P

At t=0:             a      0       (initial condition)

At t=t:            a-x      x

Say x be the amount of reactant which forms the product in time t.

So from the rate law, we have

                 rate of decomposition = k (R)ⁿ

Where k is rate constant , R is amount of reactant at time t and n is the order of the reaction

From the question, at the instant when 5% and 20% have reacted, we will be left with 95% and 80% of the reactant respectively. So writing the rate law equation:

1.07 = k ( 95a / 100)ⁿ

0.76 = k ( 80a/100)ⁿ

Dividing these two equations, we get:

(1.07 / 0.76 ) = ( 95/80 )ⁿ

Taking logarithm on both sides we get

n = ( ㏒ (1.07 / 0.76) ) ÷ ㏒(95/80) = 2.0067 ≈ 2

Therefore the reaction is of order 2

a electronegativity decreases beacuse the size of nucleus is bigger.