The decomposition of phosphine $(P H_{3})$ on tungsten at low pressure is a first-order reaction. It is because the:

rate is independent of the surface coverage

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rate of decomposition is very slow

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rate is proportional to the surface coverage

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rate is inversely proportional to the surface coverage

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Solution

The correct option is D rate is proportional to the surface coverage
The decomposition of $P H_{3}$ on tungsten at low pressure is a first order reaction because surface area coverage is proportional to partial pressure of $P H_{3}$ . Hence, rate is proportional to the surface coverage.

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The decomposition of phosphine (PH₃)on tungsten at low pressure is a first-order reaction. It is because the:

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The decomposition of phosphine (PH3) on tungsten at low pressure is a first-order reaction. It is because the:

The correct option is D rate is proportional to the surface coverageThe decomposition of PH3 on tungsten at low pressure is a first order reaction because surface area coverage is proportional to partial pressure of PH3. Hence, rate is proportional to the surface coverage.

The decomposition of phosphine $(P H_{3})$ on tungsten at low pressure is a first-order reaction. It is because the:

The correct option is D rate is proportional to the surface coverage
The decomposition of $P H_{3}$ on tungsten at low pressure is a first order reaction because surface area coverage is proportional to partial pressure of $P H_{3}$ . Hence, rate is proportional to the surface coverage.