Question

The half-period T for the decomposition of ammonia on tungsten wire was measured for different initial pressures P of ammonia at 25${}^o$C. Then:

P(mm Hg)	11	21	48	73	120
T(sec)	48	92	210	320	525

• A. Zero order reaction
• B. First order reaction
• C. Rate constant for reaction is 0.114 mol lit${}^{-1}$ sec${}^{-1}$
• D. Rate constant for reaction is 1.14 seconds

Answer 1

Answer: (A), (C)

The correct options are
A Zero order reaction
C Rate constant for reaction is 0.114 mol lit${}^{-1}$ sec${}^{-1}$

For nth ordr reaction, $t_{1/2}\alpha \left(a{)}^{1-n}or\right(1-n)=\frac{log{t}_{1/2}^{\prime }-logt{"}_{1/2}}{log{a}^{\prime }-loga"}$
In terms of partial pressures, $t_{1/2}\alpha \frac{1}{p^{n-1}}$
Substitute values in the above expression.
$\frac{21}{11}={\left(\frac{49}{92}\right)}^{n-1}n=0$
Thus the reaction is of zero order.
The expression for the rate constant is $t_{1/2}=\frac{P}{2K}$ or $K=\frac{P}{2t_{1/2}}$
$K=\frac{11}{2\times 48}=0.114sec.$.
Thus, the rate constant of the reaction is $0.114sec$