Question

Consider the reaction8H₂S(g) + 402(g) →8H₂O(g) + S8 (g)AH₂S/At = -0.021 M/s

Answer 1

`8H₂S_((g))+4O_{2(g)\text{ }}→\text{ }8H₂O_((g))+S_(8(g))`
`v=(-1)/(8)(\Delta\lbrack H_2S\rbrack)/(\Delta t)=(-1)/(4)(\Delta\lbrack O_2\rbrack)/(\Delta t)=(1)/(8)(\Delta\lbrack H_2O\rbrack)/(\Delta t)=(1)/(1)(\Delta\lbrack S_8\rbrack)/(\Delta t)`
`(-1)/(8)(\Delta\lbrack H_2S\rbrack)/(\Delta t)=(1)/(1)(\Delta\lbrack S_8\rbrack)/(\Delta t)`
`(\Delta\lbrack S_8\rbrack)/(\Delta t)=\text{ }(-1)/(8)(\Delta\lbrack H_2S\rbrack)/(\Delta t)=(-1)/(8)(-0.021\text{ M/s\rparen = 0.0026 M/s}`
`v=(\Delta\lbrack S_8\rbrack)/(\Delta t)=\text{ 0.0026 M/s}`

Part C the answer is Δ[S8]/Δt = 0.0026 M/s

Parte D the answer is v = 0.0026 M/s