The rate constant for the decomposition of N_{2}O_{5} at various temperatures is given below:

T/°C | 0 | 20 | 40 | 60 | 80 |

10 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |

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.

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 = (y_{2}-y_{1})/(x_{2}-x_{1})

By substituting the values, slope = -12.301

⇒ –E_{a}/R = -12.301

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

⇒ E_{a} = 8.314 JK^{-1}mol^{-1} × 12.301 K

⇒ E_{a} = 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 = 2.91×10^{6}

When T = 30^{0}C, hence T = 30 + 273 = 303K

⇒

⇒

⇒ ln k = -2.8

⇒ k = 6.08x10^{-2}s^{-1}.

When T = 50^{0}C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 × 10^{6}, E_{a} = 102.27 kJ mol^{-1}

Ln K = ln(2.91 × 10^{6})- 102.27/(8.314 × 323)

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

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