My answer to the question is "Gas particles are acting like tiny,solid spheres".
Answer : The rate constant at 785.0 K is, 
Explanation :
According to the Arrhenius equation,

or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= 
= rate constant at
= ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = 
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)

Therefore, the rate constant at 785.0 K is, 
There are 2 significant figures. All numbers in a whole number are significant.
The question is incomplete. The complete question is :
Hydrogen is manufactured on an industrial scale by this sequence of reactions:


The net reaction is :

Write an equation that gives the overall equilibrium constant
in terms of the equilibrium constants
and
. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.
Solution :

...............(1)

...................(2)

![$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
On multiplication of equation (1) and (2), we get
![$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%5Ctimes%20K_2%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5Ctimes%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
.................(4)
Comparing equation (3) and equation (4), we get
