4.31 The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and Ea . Predict the rate constant at 30° and 50°C. 

 

4.31 To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^-Ea/RT

Where, k- Rate constant

A- Constant

E_a-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A- (E_a/RT). equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.

Slope = (y₂-y₁)/ (x₂-x₁)

By substituting the values, slope = -12.301

? –E_a/R = -12.301

But, R = 8.314 JK^-1mol^-1

? _aE= 8.314 JK^-1mol^-1 * 12.301 K

? _aE= 102.27 kJ mol^-1

Substituting the values in equation 1 for data at T = 273K

(? At T = 273K, ln k =-7.147)

On solving, we get ln A = 37.

...more

4.31 To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^-Ea/RT

Where, k- Rate constant

A- Constant

E_a-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A- (E_a/RT). equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.

Slope = (y₂-y₁)/ (x₂-x₁)

By substituting the values, slope = -12.301

? –E_a/R = -12.301

But, R = 8.314 JK^-1mol^-1

? _aE= 8.314 JK^-1mol^-1 * 12.301 K

? _aE= 102.27 kJ mol^-1

Substituting the values in equation 1 for data at T = 273K

(? At T = 273K, ln k =-7.147)

On solving, we get ln A = 37.911

? A = 2.91*10⁶

When T = 30⁰C, hence T = 30 + 273 = 303K

? ln k =-2.8

? k = 6.08x10^-2s^-1.

When T = 50⁰C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 * 10⁶, E_a = 102.27 kJ mol^-1

Ln K = ln (2.91 * 10⁶)- 102.27/ (8.314 * 323)

Thus, ln k = -0.5 and k = 0.607s^-1.

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<p><strong>4.31&nbsp;</strong>To convert the temperature in °C to °K we add 273 K.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690294854phpXp6pD1_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690294854phpXp6pD1.jpeg" alt="" width="406" height="425"></picture></div></div><p>The graph is given as:</p><p>The Arrhenius equation is given by k = Ae^-Ea/RT</p><p>Where, k- Rate constant</p><p>A- Constant</p><p>E_a-Activation Energy</p><p>R- Gas constant</p><p>T-Temperature</p><p>Taking natural log on both sides, </p><p>ln k = ln A- (E_a/RT). equation 1</p><p>By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.</p><p>Slope = (y₂-y₁)/ (x₂-x₁)</p><p>By substituting the values, slope = -12.301</p><p>? –E_a/R = -12.301</p><p>But, R = 8.314 JK^-1mol^-1</p><p>? _aE= 8.314 JK^-1mol^-1 * 12.301 K</p><p>? _aE= 102.27 kJ mol^-1</p><p>Substituting the values in equation 1 for data at T = 273K</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690294992phpaV7utQ_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690294992phpaV7utQ.jpeg" alt="" width="206" height="73"></picture></div></div><p> (? At T = 273K, ln k =-7.147)</p><p>On solving, we get ln A = 37.911</p><p>? A = 2.91*10⁶</p><p>When T = 30⁰C, hence T = 30 + 273 = 303K</p><p>? ln k =-2.8</p><p>? k = 6.08x10^-2s^-1.</p><p>When T = 50⁰C, hence T = 50 + 273 = 323K</p><p>Substituting in equation 1, </p><p>A = 2.91 * 10⁶, E_a = 102.27 kJ mol^-1</p><p>Ln K = ln (2.91 * 10⁶)- 102.27/ (8.314 * 323)</p><p>Thus, ln k = -0.5 and k = 0.607s^-1.</p>

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