In the reaction between NO and $\mathrm{H_2}$ the following data are obtainedExperiment I: $P_{\mathrm{H_2}}$ constant
$P_{\mathrm{NO}}$ (mm of Hg)
359
300
152
$-\dfrac{dP_{\mathrm{NO}}}{dt}$
1.50
1.03
0.25
Experiment II : $P_{\mathrm{NO}}$ = constant
$P_{\mathrm{H_2}}$ (mm of Hg)
289
205
147
$-\dfrac{dP_{\mathrm{H_2}}}{dt}$
1.60
1.10
0.79
The orders with respect to $\mathrm{H_2}$ and NO are
(a)1 with respect to NO and 2 with respect to $\mathrm{H}_{2}$
(b)2 with respect to NO and 1 with respect to $\mathrm{H}_{2}$
(c)1 with respect to NO and 3 with respect to $\mathrm{H}_{2}$
(d)2 with respect to NO and 2 with respect to $\mathrm{H}_{2}$
Answer
Answer: B ✓ checked by 4AB · confidence medium
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Experiment I (P(H₂) constant): from 359 to 152 mm, P(NO) falls by a factor of 2.36 while the rate falls by 1.50/0.25 = 6. Order n = ln 6 / ln 2.36 = 1.79/0.86 ≈ 2. The pair 359/300 gives the same: ln(1.456)/ln(1.197) ≈ 2.1.
Experiment II (P(NO) constant): from 289 to 147 mm, P(H₂) falls by a factor of 1.97 while the rate falls by 1.60/0.79 = 2.03. Order m = ln 2.03 / ln 1.97 ≈ 1.
Rate = k[NO]²[H₂]: second order in NO and first order in H₂. Answer (b).