Answer:
First-order with respect to hydrogen.
Explanation:
Hello!
In this case, considering that the rate law of this reaction can be expressed via:
[tex]r=k*p_{H_2}^m*p_{ICl}^n[/tex]
If we want to know m, the order of reaction with respect to hydrogen, we need to relate the experiments 1 and 3 in order to get rid of the pressure of ICl:
[tex]\frac{r_1}{r_3} =\frac{k*p_{H_2,1}^m*p_{ICl,1}^n}{k*p_{H_2,3}^m*p_{ICl,3}^n}[/tex]
Thus, we plug in the given rates, and pressures to get:
[tex]\frac{1.34}{0.266} =\frac{k*250^m*325^n}{k*50^m*325^n}[/tex]
So we can cancel wout k and 325^n:
[tex]\frac{1.34}{0.266} =\frac{250^m}{50^m}[/tex]
Next we solve for m, the order of reaction with respect to hydrogen:
[tex]5.04 =5^m\\\\log(5.04)=m*log(5)\\\\m=\frac{log(5.04)}{log(5)} \\\\m=1.0[/tex]
It means it is first-order with respect to hydrogen.
Regards!
