Reaction rate Determine the reaction rate at low pressure for the mechanism reaction H + H + M → H2 + M at temperature of 3,000 k, where M is H2 as a function of [H].

Question

Reaction rate
Determine the reaction rate at low pressure for the mechanism reaction

[latex]H + H + M → H_{2} + M[/latex]

at temperature of 3,000 k, where M is [latex]H_{2}[/latex] as a function of [H].

Accepted Answer

According to Eq. (3.22), the chemical reaction rate for the reaction

[latex]r_{A}=\frac{d\left[A\right]}{dt}=-k\left[A\right]^{a }\left[B\right]^{b}[/latex] (3.22)

H + H + M → M + [latex]H_{2}[/latex]

is described in terms of the formation rate of [latex]H_{2}[/latex]

[latex]\frac{d\left[H_{2}\right] }{dt}=k\left[H\right]^{2}[/latex]

where the reaction rate constant k can be described using Eq. (3.25)

[latex]k=AT^{B} \exp\left(-\frac{E_{A}}{RT} \right)[/latex] (3.25)

Therefore, the chemical reaction rate is described as

[latex]\frac{d\left[H_{2}\right] }{dt}=AT^{B} \exp\left(-\frac{E_{A}}{RT} \right)[H]^{2}[/latex]

There is no negative sign because [latex]H_{2}[/latex] is produced rather than consumed in this one-way reaction.

From Table 3.1, we can get

A = 0.7 × [latex]10^{16}[/latex], B = -0.6, and [latex]E_{A}[/latex] = 0

Thus the rate of reaction at T = 3,000 K is

[latex]\frac{d\left[H_{2}\right] }{dt}= 0.7 × 10^{16} ×3,000^{-0.6} \exp(-0)[H]^{2} = 5.74 × 10^{19}[H]^{2}[/latex]

The solution indicates that molecular hydrogen is produced extremely fast at high temperature.

Table 3.1 Rate coefficients for H–O reactions

Reaction	A[latex]\left[(cm^{3}/mole)^{n-1}\right]^{n}[/latex]	B	[latex]E_{A}[/latex](J/mole)	Temperature Range (K)
H + [latex]O_{2}[/latex] → OH + O	1.2 × [latex]10^{17}[/latex]	−0.91	69.1	300–2,500
OH + O → [latex]O_{2}[/latex]+ H	1.8 × [latex]10^{13}[/latex]	0	0	300–2,500
O + [latex]H_{2}[/latex] → OH + H	1.5 × [latex]10^{7}[/latex]	2.0	31.6	300–2,500
[latex]OH + H_{2} → H_{2}O + H[/latex]	1.5 × [latex]10^{8}[/latex]	1.6	13.8	300–2,500
[latex]H + H_{2}O → OH + H_{2}[/latex]	4.6 × [latex]10^{8}[/latex]	1.6	77.7	300–2,500
O + [latex]H_{2}O[/latex] → OH + OH	1.5 × [latex]10^{10}[/latex]	1.14	72.2	300–2,500
H + H + M → [latex]H_{2}[/latex] + M
M = Ar (low P)	6.4 × [latex]10^{17}[/latex]	−1.0	0	300–5,000
M = [latex]H_{2}[/latex] (low P)	0.7 × [latex]10^{16}[/latex]	-0.6	0	100–5,000
[latex]H_{2}[/latex] + M → H + H + M
M = Ar (low P)	2.2 × [latex]10^{14}[/latex]	0	402	2,500–8,000
[latex]M + H_{2}[/latex] (low P)	8.8 × [latex]10^{14}[/latex]	0	402	2,500–8,000
H + OH + M → [latex]H_{2}O[/latex] + M
M = [latex]H_{2}O[/latex] (low P)	1.4	−2.0	0	1,000–3,000
[latex]H_{2}O[/latex] + M → H + OH + M
M = [latex]H_{2}O[/latex] (low P)	1.6 × [latex]10^{17}[/latex]	0	478	2,000–5,000
O + O + M → [latex]O_{2}[/latex] + M
M = Ar (low P)	1.0 × [latex]10^{17}[/latex]	−1.0	0	300–5,000
[latex]O_{2}[/latex] + M → O + O + M
M = Ar (low P)	1.2 × [latex]10^{14}[/latex]	0	451	2,000–10,000

Question 3.5: Reaction rate Determine the reaction rate at low pressure fo...