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Snowcat [4.5K]
3 years ago
7

Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.44 × 1

0-2 M/s, what is the rate of formation of NOCl?2 NO(g) + Cl2(g) → 2 NOCl(g)1.11 × 10-1 M/s2.22 × 10-2 M/s8.88 × 10-2 M/s1.61 × 10-2 M/s4.44 × 10-2 M/s
Chemistry
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

8.88 x 10⁻² M/s

Explanation:

The rate  of reaction for:

                    NO(g) + Cl₂ (g) ⇒  2NOCl(g)

is rate = -ΔNO/Δt = -ΔCl2/Δt = 1/2 ΔNOCl/Δt

so  ΔNOCl/Δt = 2 ΔCl2/Δt  = 2  x 4.44 × 10⁻²  M/s = 8.88 x 10⁻² M/s

In general given a reaction

                            aA + bB ⇒ cC + dD

rate = -1/a ΔA/Δt = -1/b ΔB/Δt = 1/c ΔC/Δt = 1/d ΔD/Δt

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Answer:  ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)

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\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]

\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]

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Now put all the given values in this expression, we get

\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]

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