The reaction 2NO + 2H_(2) to N_(2) +2H_(2)O follows the mechanism: I) NO + NO_(2) ltimplies N_(2)O_(2) (fast) II) N_(2)O_(2) + H_(2) to N_(2)O + H_(2)O (slow) III) N_(2)O + H_(2) to N_(2) + H_(2)O (fast) the rate constant for the slow step (II) is 1.2 xx 10^(-4) mol^(-1)Lmin^(-1) while the equilibrium constant for step (I) is 1.4 xx 10^(-2) what is the rate of reaction when the concentration of NO and H_(2) each is 0.5 mol^(-1)L?

The reaction 2NO + 2H_(2) to N_(2) +2H_(2)O follows the mechanism: I)

1.	The reaction 2NO + 2H_(2) to N_(2) +2H_(2)O follows the mechanism: I) NO + NO_(2) ltimplies N_(2)O_(2) (fast) II) N_(2)O_(2) + H_(2) to N_(2)O + H_(2)O (slow) III) N_(2)O + H_(2) to N_(2) + H_(2)O (fast) the rate constant for the slow step (II) is 1.2 xx 10^(-4) mol^(-1)Lmin^(-1) while the equilibrium constant for step (I) is 1.4 xx 10^(-2) what is the rate of reaction when the concentration of NO and H_(2) each is 0.5 mol^(-1)L?
Answer» `2.1 xx 10^(-7) mol L^(-1)min^(-1)` `3.2 xx 10^(-6)mol L^(-1)min^(-1)` `3.5 xx 10^(-4)mol L^(-1)min^(-1)` None of the above. Solution :`k_("obs") = k xx k_(c) = 1.2 xx 10^(-4)` `=1.2 xx 10^(-4)mol^(-1)L min^(-1) xx (1.4 xx 10^(-2))` `=1.68 xx 10^(-6) mol^(-1)L min^(-1)` RATE =`k_("obs")[NO]^(2)[H_(2)]` `=(1.68 xx 10^(-6) mol^(-1)L min^(-1)) xx (0.5 molL^(-1))^(3)`