At 300 K,for the reaction A rightarrow P,the | vedclass

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(A) The entropy change of the system is given by the formula: $\Delta S_{sys} = \frac{q_{rev}}{T}$.Assuming the process is reversible,the heat absorbe

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$1.5$

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(A) The entropy change of the system is given by the formula: $\Delta S_{sys} = \frac{q_{rev}}{T}$.Assuming the process is reversible,the heat absorbed $q_{sys}$ is calculated as:$q_{sys} = \Delta S_{sys} 	imes T$$q_{sys} = 5 \ J \ K^{-1} \ mol^{-1} 	imes 300 \ K = 1500 \ J \ mol^{-1}$.To convert the value into $kJ \ mol^{-1}$,divide by $1000$:$q_{sys} = \frac{1500}{1000} \ kJ \ mol^{-1} = 1.5 \ kJ \ mol^{-1}$.

At $300 \ K$,for the reaction $A \rightarrow P$,the $\Delta S_{sys}$ is $5 \ J \ K^{-1} \ mol^{-1}$. What is the heat absorbed (in $kJ \ mol^{-1}$) by the system?

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