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Question: For the reaction \(N_{2} + 3H_{2}\)⇌\(2NH_{3}\), if \(\frac{\Delta\lbrack NH_{3}\rbrack}{\Delta t} ...

For the reaction N2+3H2N_{2} + 3H_{2}⇌2NH32NH_{3}, if

Δ[NH3]Δt=2×10−4mol 6mul−1s−1,\frac{\Delta\lbrack NH_{3}\rbrack}{\Delta t} = 2 \times 10^{- 4}mol\mspace{6mu} l^{- 1}s^{- 1}, the value of −Δ[H2]Δt\frac{- \Delta\lbrack H_{2}\rbrack}{\Delta t} would

be

A

1×10−4moll−1s−11 \times 10^{- 4}moll^{- 1}s^{- 1}

B

3×10−4moll−1s−13 \times 10^{- 4}moll^{- 1}s^{- 1}

C

4×10−4moll−1s−14 \times 10^{- 4}moll^{- 1}s^{- 1}

D

6×10−4moll−1s−16 \times 10^{- 4}moll^{- 1}s^{- 1}

Answer

3×10−4moll−1s−13 \times 10^{- 4}moll^{- 1}s^{- 1}

Explanation

Solution

N2+3H2N_{2} + 3H_{2}⇌ 2NH32NH_{3}

−Δ[N2]Δt=−13Δ[H2]Δt=12Δ[NH3]Δt\frac{\mathbf{- \Delta\lbrack}\mathbf{N}_{\mathbf{2}}\mathbf{\rbrack}}{\mathbf{\Delta t}}\mathbf{= -}\frac{\mathbf{1}}{\mathbf{3}}\frac{\mathbf{\Delta\lbrack}\mathbf{H}_{\mathbf{2}}\mathbf{\rbrack}}{\mathbf{\Delta t}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{\Delta\lbrack N}\mathbf{H}_{\mathbf{3}}\mathbf{\rbrack}}{\mathbf{\Delta t}}; ∴Δ[H2]Δt=32×Δ[NH3]Δt=32×2×10−4\mathbf{\therefore}\frac{\mathbf{\Delta}\mathbf{\lbrack}\mathbf{H}_{\mathbf{2}}\mathbf{\rbrack}}{\mathbf{\Delta}\mathbf{t}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{\times}\frac{\mathbf{\Delta}\mathbf{\lbrack N}\mathbf{H}_{\mathbf{3}}\mathbf{\rbrack}}{\mathbf{\Delta}\mathbf{t}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{\times}\mathbf{2}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{-}\mathbf{4}}

=3×10−4moll−1s−1\mathbf{= 3}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{-}\mathbf{4}}\mathbf{mol}\mathbf{l}^{\mathbf{-}\mathbf{1}}\mathbf{s}^{\mathbf{-}\mathbf{1}}.