The reaction below takes place in a constant pressure kilometer heat flows from the reaction into the kilometer create an equation that relates to the heat of the kilometer to the heat of reaction
Added by Christopher D.
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In a constant pressure calorimetry setup, we have a chemical reaction occurring in a calorimeter, where heat is exchanged between the reaction and the calorimeter. Show more…
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1. For the reaction 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) ΔH° = -369 kJ and ΔS° = -15.3 J/K The equilibrium constant, K, would be greater than 1 at temperatures _______ above/below Kelvin. Select above or below in the first box and enter the temperature in the second box. Assume that ΔH° and ΔS° are constant. 2. For the reaction CO(g) + Cl2(g) → COCl2(g) ΔH° = -108 kJ and ΔS° = -137 J/K The equilibrium constant, K, would be greater than 1 at temperatures _______ above/below Kelvin. Select above or below in the first box and enter the temperature in the second box. Assume that ΔH° and ΔS° are constant.
Shaiju T.
(A): In the following reaction $$ \begin{aligned} &\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) \\ &\Delta \mathrm{H}=\Delta \mathrm{E}-\mathrm{RT} \end{aligned} $$ $(\mathbf{R}): \Delta \mathrm{H}$ is related to $\Delta \mathrm{E}$ by equation $$ \Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} $$
For a certain reaction, ΔH° = - 83.5 kJ and ΔS° = -148 J/K. If the values of ΔH° and ΔS° can be assumed to be constant for all temperatures, the temperature at which the reaction is at equilibrium is ΔG = ΔH - TΔS
Stephen P.
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