Question

The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K. Calculate the standard change in enthalpy for this reaction (assuming ∆H° is temperature independent).

Answer 1

Answer:

$\text { The standard change in enthalpy for this reaction is } \Delta \mathrm{H}=2.397 \times 10^{4} \mathrm{J} / \mathrm{mol}$

Explanation:

Let’s assume$\text { At } 300 \mathrm{K}, \mathrm{k}_{\mathrm{eq}}=\mathrm{x}$

Thus as per given information

$\text { At } 350 \mathrm{K}, \mathrm{k}_{\mathrm{eq}}=3.95 \mathrm{x}$

As we know:

$\ln \left(\frac{k_{2}}{k_{1}}\right)=-\frac{\Delta \mathrm{H}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$

$\ln \left(\frac{3.95 x}{x}\right)=\frac{-\Delta \mathrm{H}}{8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}$

$8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \times \ln 3.95=-\Delta \mathrm{H} \times\left(-4.761 \times 10^{-4} \mathrm{K}^{-1}\right)$

$\Delta \mathrm{H}=\frac{8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \times 1.373}{-4.761 \times 10^{-4} \mathrm{K}^{-1}}$

$\Delta \mathrm{H}=2.397 \times 10^{4} \mathrm{J} / \mathrm{mol}$

Answer 2

The standard change in enthalpy for this reaction is mathematically given as

dH=2.397 *10^{4} J/mol

What is the standard change in enthalpy for this reaction?

Question Parameter(s):

The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K.

Generally, the equation for the change in enthalpy   is mathematically given as

$\ln \ (\frac{k_{2}}{k_{1}})=-\frac{\Delta \mathrm{H}}{R} (\frac{1}{T_{2}}-\frac{1}{T_{1}})$

Therefore

$8.314 {J} {mol}^{-1} {K}^{-1} *\ln 3.95=-d{H} *(-4.761 \times 10^{-4} {K}^{-1})$

dH=2.397 *10^{4} J/mol

In conclusion

dH=2.397 *10^{4} J/mol

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