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The decomposition of NH 3 on platinum surface is zero order. What are the rates of production of N 2and H 2 if 4 1 1 2.5 10 k mol Ls - - -= × ?

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12 Years agoGrade upto college level
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ApprovedApproved Tutor Answer1 Year ago

To tackle the question regarding the decomposition of ammonia (NH₃) on a platinum surface, we first need to understand the implications of a zero-order reaction. In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. This means that the rate of production of products is solely determined by the rate constant.

Understanding the Reaction

The decomposition of ammonia can be represented by the following balanced chemical equation:

  • 2 NH₃(g) → N₂(g) + 3 H₂(g)

From this equation, we can see that for every 2 moles of ammonia that decompose, 1 mole of nitrogen (N₂) and 3 moles of hydrogen (H₂) are produced.

Rate Constant and Reaction Rate

Given that the rate constant (k) for the reaction is 4.1 × 10-2 mol L-1 s-1, we can directly use this value to determine the rates of production of N₂ and H₂. Since the reaction is zero-order, the rate of production of products is equal to the rate constant.

Calculating Production Rates

Since the rate of the reaction is constant, we can express the rates of production of N₂ and H₂ as follows:

  • Rate of production of N₂ = k / 2
  • Rate of production of H₂ = (3/2) * (k / 2)

Now, substituting the value of k into these equations:

  • Rate of production of N₂ = (4.1 × 10-2 mol L-1 s-1) / 2 = 2.05 × 10-2 mol L-1 s-1
  • Rate of production of H₂ = (3/2) * (2.05 × 10-2 mol L-1 s-1) = 3.075 × 10-2 mol L-1 s-1

Final Results

To summarize, the rates of production for the decomposition of ammonia on a platinum surface are:

  • Rate of production of N₂: 2.05 × 10-2 mol L-1 s-1
  • Rate of production of H₂: 3.075 × 10-2 mol L-1 s-1

This analysis illustrates how the zero-order kinetics simplifies the calculations, allowing us to directly relate the rate constant to the production rates of the products formed during the reaction. If you have any further questions about reaction kinetics or related topics, feel free to ask!