Respuesta :

Answer:

[tex]\frac{[\Delta O_2]}{\Delta t} = -0.0165\ M[/tex]

[tex] \frac{[\Delta H_2O]}{\Delta t}= 0.033\ M/s[/tex]

[tex]\frac{[\Delta S_8]}{\Delta t} = 0.004125\ M/s[/tex]

Rate of the reaction = 0.004125 M/s

Explanation:

For a general reaction,

[tex]aA + bB \rightarrow cC + dD[/tex]

Rate of the reaction = [tex]-\frac{1}{a}\times \frac{\Delta A}{\Delta t} =-\frac{1}{b}\times \frac{\Delta B}{\Delta t}= \frac{1}{c}\times \frac{\Delta C}{\Delta t} = \frac{1}{d}\times \frac{\Delta D}{\Delta t}[/tex]

So, for the given reaction,

[tex]8H_2S(g) + 4O_2(g) \rightarrow 8H_2O(g) + S_8(g)[/tex]

Rate of the reaction = [tex]-\frac{1}{8}\times \frac{\Delta H_2S}{\Delta t} =-\frac{1}{4}\times \frac{\Delta O_2}{\Delta t}= \frac{1}{8}\times \frac{\Delta H_2O}{\Delta t} = \frac{1}{1}\times \frac{\Delta S_8}{\Delta t}[/tex]

[tex]\frac{\Delta H_2S}{\Delta t} = -0.033\ M/s[/tex]

Calculation of [ΔO2]/Δt

[tex]-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t} =-\frac{1}{4}\times \frac{[\Delta O_2]}{\Delta t}[/tex]

[tex]-\frac{1}{8}\times (-0.33) =-\frac{1}{4}\times \frac{[\Delta O_2]}{\Delta t}[/tex]

[tex]-\frac{[\Delta O_2]}{\Delta t} = \frac{4}{8} \times (0.033) = 0.0165\ M[/tex]

[tex]\frac{[\Delta O_2]}{\Delta t} = -0.0165\ M/s[/tex]

Calculation of [ΔH2O]/Δt

[tex]-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t} =\frac{1}{8}\times \frac{[\Delta H_2O]}{\Delta t}[/tex]

[tex] \frac{[\Delta H_2O]}{\Delta t}=-\frac{8}{8}\times \frac{[\Delta H_2S]}{\Delta t}[/tex]

[tex] \frac{[\Delta H_2O]}{\Delta t}=-\frac{8}{8}\times (-0.033)[/tex]

[tex] \frac{[\Delta H_2O]}{\Delta t}= 0.033\ M/s[/tex]

Calculation of [ΔS8]/Δt

[tex]-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t} =\frac{1}{1}\times \frac{[\Delta S_8]}{\Delta t}[/tex]

[tex]\frac{[\Delta S_8]}{\Delta t}=-\frac{1}{8}\times (-0.033)=0.004125\ M/s[/tex]

Calculation of Rate of Reaction

Rate of the reaction = [tex]-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t}[/tex]

Rate of the reaction =[tex]-\frac{1}{8}\times (-0.033)=0.004125\ M/s[/tex]