Answer:
\(\frac{[\Delta O_2]}{\Delta t} = -0.0165\ M\)
\(\frac{[\Delta H_2O]}{\Delta t}= 0.033\ M/s\)
\(\frac{[\Delta S_8]}{\Delta t} = 0.004125\ M/s\)
Rate of the reaction = 0.004125 M/s
Explanation:
For a general reaction,
\(aA + bB \rightarrow cC + dD\)
Rate of the reaction = \(-\frac{1}{a}\times \frac{\Delta A}{\Delta t} =-\frac{1}{b}\times \frac{\Delta B}{\Delta t}= \frac{1}{c}\times \frac{\Delta C}{\Delta t} = \frac{1}{d}\times \frac{\Delta D}{\Delta t}\)
So, for the given reaction,
\(8H_2S(g) + 4O_2(g) \rightarrow 8H_2O(g) + S_8(g)\)
Rate of the reaction = \(-\frac{1}{8}\times \frac{\Delta H_2S}{\Delta t} =-\frac{1}{4}\times \frac{\Delta O_2}{\Delta t}= \frac{1}{8}\times \frac{\Delta H_2O}{\Delta t} = \frac{1}{1}\times \frac{\Delta S_8}{\Delta t}\)
\(\frac{\Delta H_2S}{\Delta t} = -0.033\ M/s\)
Calculation of [ΔO2]/Δt
\(-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t} =-\frac{1}{4}\times \frac{[\Delta O_2]}{\Delta t}\)
\(-\frac{1}{8}\times (-0.33) =-\frac{1}{4}\times \frac{[\Delta O_2]}{\Delta t}\)
\(-\frac{[\Delta O_2]}{\Delta t} = \frac{4}{8} \times (0.033) = 0.0165\ M\)
\(\frac{[\Delta O_2]}{\Delta t} = -0.0165\ M/s\)
Calculation of [ΔH2O]/Δt
\(-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t} =\frac{1}{8}\times \frac{[\Delta H_2O]}{\Delta t}\)
\(\frac{[\Delta H_2O]}{\Delta t}=-\frac{8}{8}\times \frac{[\Delta H_2S]}{\Delta t}\)
\(\frac{[\Delta H_2O]}{\Delta t}=-\frac{8}{8}\times (-0.033)\)
\(\frac{[\Delta H_2O]}{\Delta t}= 0.033\ M/s\)
Calculation of [ΔS8]/Δt
\(-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t} =\frac{1}{1}\times \frac{[\Delta S_8]}{\Delta t}\)
\(\frac{[\Delta S_8]}{\Delta t}=-\frac{1}{8}\times (-0.033)=0.004125\ M/s\)
Calculation of Rate of Reaction
Rate of the reaction = \(-\frac{1}{8}\times \frac{[\Delta H_2S]}{\Delta t}\)
Rate of the reaction =\(-\frac{1}{8}\times (-0.033)=0.004125\ M/s\)
