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NH4+(aq) + NO2−(aq) → N2(g) + 2H2O(l) is given by rate = k[NH4+][NO2−]. At a certain temperature, the rate constant is 4.10 × 10−4 /M·s. Calculate the rate of the reaction at that temperature if [NH4+] = 0.301 M and [NO2−] = 0.160 M.

Respuesta :

Answer:

[tex]rate = 1.97\times 10^{-5}\ M/s[/tex]

Explanation:

Given that:-

[tex]rate = k[NH_4^+][NO_2^-][/tex]

k = [tex]4.10\times 10^{-4}\ /Ms[/tex]

[tex][NH_4^+]=0.301\ M[/tex]

[tex][NO_2^-]=0.160\ M[/tex]

So,

[tex]rate = 4.10\times 10^{-4}\times 0.301\times 0.160\ M/s=1.97\times 10^{-5}\ M/s[/tex]

[tex]1.97\times 10^{-5}\ M/s[/tex] is the rate of the reaction at that temperature if [NH4+] = 0.301 M and [NO2−] = 0.160 M.

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