Question

The pressure of a gas decomposing at the surface of a solid catalyst has been measured at different times and the results are given below
t(sec) 0 100 200 300
Pr.(Pascal) $4\times {10}^3$ $3.5\times {10}^3$ $3\times {10}^3$ $2.5\times {10}^3$
The order of reaction and its rate constant is:

• A. 0, K=5 pa/s
• B. 1, K=6 pa/s
• C. 0, K=6 pa/s
• D. 1, K=5 pa/s

Answer 1

The correct option is A 0, K=5 pa/s

For every 100 sec, the pressure decreases by $0.5\times {10}^3$
Hence, the change in pressure is independent of the initial pressure and the reaction is of zero order.
$P-P_0=-kt$
for $t=100$
$3.5\times {10}^3-4.0\times {10}^3=-k\times 100$
$k=5pa/s$