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Chemistry

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____ [38]3 years ago

6 0

Answer:

Explanation:

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Calculate δs∘rxn for the reaction2no(g) o2(g)→2no2(g)express your answer to one decimal place and include the appropriate units

Anna35 [415]

The δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.

Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit $JK^{-1} mol^{-1}$ .

Entropy of the reaction can be calculated by the reaction.

Δ $S^{0} rxn$ = 2 mol × $S^{0} (NO_{2} (g) - 2 mol$ × $S^{0} NO (g)$ - 1 mol × $S^{0} (O_{2} )$

Δ $S^{0} rxn$ = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol ×205.2 J/mol.K

Δ $S^{0} rxn$ = -146.8 J/K

Therefore, the δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

To know more about reaction

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4 0

1 year ago

start out as one kind of rock, but large amounts of pressure and heat change them into a different kind.

Eddi Din [679]

Answer:

Metamorphic

Explanation:

6 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$