The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from 300.0 \(\mathrm{K}\) to 350.0 \(\mathrm{K}\) . Calculate the standard change in enthalpy $\left(\Delta H^{\circ}\right)\( for this reaction (assuming \)\Delta H^{\circ}$ is temperature-independent).

We need to find the standard change in enthalpy, and one equation relating it to temperature and the equilibrium constant is the Van't Hoff equation: \[\frac{d\ln K}{dT} = \frac{\Delta H^{\circ}}{R \cdot T^2}\] where K is the equilibrium constant, T is the temperature in Kelvin, R is the ideal gas constant, and ΔH° is the standard enthalpy change. However, as we are given the initial and final temperatures and the factor by which the equilibrium constant increases, we can rearrange the Van't Hoff equation into the following form: \[\ln \frac{K_2}{K_1} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Here, \(K_1\) and \(K_2\) are the initial and final equilibrium constants, respectively, and \(T_1\) and \(T_2\) are the initial and final temperature values.

We know that \(K_2 = 6.67 K_1\), \(T_1 = 300.0\) K and \(T_2 = 350.0\) K. First, let's substitute the given values into the equation: \[\ln \frac{6.67 K_1}{K_1} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{350.0} - \frac{1}{300.0}\right)\]

Now, let's simplify the equation and solve for ΔH°: \[\ln 6.67 = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{350.0} - \frac{1}{300.0}\right)\] \[\Delta H^{\circ} = -R\left(\frac{1}{350.0} - \frac{1}{300.0}\right)^{-1} \ln 6.67\] We can now plug in the value of R, the ideal gas constant, which is \(8.314 \text{ J/mol K}\): \[\Delta H^{\circ} = -8.314 \left(\frac{1}{350.0} - \frac{1}{300.0}\right)^{-1} \ln 6.67\]

Finally, calculate the value of ΔH°: \[\Delta H^{\circ} = -8.314 \cdot \frac{1}{\left(\frac{1}{350.0} - \frac{1}{300.0}\right)} \cdot \ln 6.67\] \[\Delta H^{\circ} \approx 9275.7 \text{ J/mol} \] So, the standard change in enthalpy for this reaction is approximately \(9275.7 \text{ J/mol}\).