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DaniilM [7]
3 years ago
10

A decrease is the concentration of nitrogen monoxide (blank#1) between NO and H2 molecules. The rate of the forward reaction the

n (blank#2).
options #1 -

stabilizes the forward reaction

increases the energy

decreases the collisions

options #2 -

remains constant

increases

decreases
Chemistry
1 answer:
OverLord2011 [107]3 years ago
5 0

In the given situation, the reaction is-

NO + H2 ↔ Products

The rate of the reaction can be expressed (in terms of the decrease in the concentration of the reactants) as-

Rate = -dΔ[NO]/dt = -dΔ[H2]/dt

Now, if the concentration of NO is decreased there will be fewer molecules of the reactant NO which would decrease the its collision with H2. As a result the rate of the forward reaction would also decrease.

Ans) A decrease in the concentration of nitrogen monoxide decreases the collisions between NO and H2 molecules. the rate of the forward reaction then decreases.

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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
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Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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