When ammonia is treated with O 2 at elevated temperatures- the rate of disappearance of ammonia is found to be 3-5- 10 -2 mol dm -3 s -1 during a measured time interval- Calculate the rate of appearance of nitric oxide and water-

Consider the following equation: 4N H _ 3 +5 O _ 2 → 4NO+6 H _ 2 O The equality in this case is 1 4 Δ[ N H _ 3 ] #160; dt = 1 4 Δ[ NO ] #160; ...

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Standard XII

Chemistry

Reaction Quotient

When ammonia ...

Question

When ammonia is treated with $O_{2}$ at elevated temperatures, the rate of disappearance of ammonia is found to be $3.5 \times 10^{- 2}$ $m o l$ ${d m}^{- 3} s^{- 1}$ during a measured time interval. Calculate the rate of appearance of nitric oxide and water.

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Solution

Consider the following equation:
$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$
The equality in this case is
$- \frac{1}{4} \frac{Δ [N H_{3}]}{d t} = \frac{1}{4} \frac{Δ [N O]}{d t} = \frac{1}{6} \frac{Δ [H_{2} O]}{d t}$

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Q. The reaction between

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Q. In the first step of the Ostwald process for synthesis of nitric acid, ammonia is converted into nitric oxide by high-temperature reaction.

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(a) How is the rate of disappearance of

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Q. Ammonia and oxygen reacts at higher temperatures as

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1.08 \times 10^{- 2} m o l l i t r e^{- 1}

3

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Q. In a catalytic reaction involving the formation of ammonia by Habers process

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When ammonia is treated with O2 at elevated temperatures, the rate of disappearance of ammonia is found to be 3.5×10−2 mol dm−3s−1 during a measured time interval. Calculate the rate of appearance of nitric oxide and water.

Consider the following equation:4NH3+5O2→4NO+6H2OThe equality in this case is−14Δ[NH3]dt=14Δ[NO]dt=16Δ[H2O]dt

When ammonia is treated with $O_{2}$ at elevated temperatures, the rate of disappearance of ammonia is found to be $3.5 \times 10^{- 2}$ $m o l$ ${d m}^{- 3} s^{- 1}$ during a measured time interval. Calculate the rate of appearance of nitric oxide and water.

Consider the following equation:
$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$
The equality in this case is
$- \frac{1}{4} \frac{Δ [N H_{3}]}{d t} = \frac{1}{4} \frac{Δ [N O]}{d t} = \frac{1}{6} \frac{Δ [H_{2} O]}{d t}$