2MnO4^- + 16H^+ + 10Cl^- → 2Mn^2+ + 5(g) + 8. Above reaction is endothermic and hence the actual temperature of the reaction vessel (isolated from the surrounding) may be different from that expected.

Question

2MnO4^- + 16H^+ + 10Cl^- → 2Mn^2+ + 5$Cl_2$(g) + 8$H_2O$. Above reaction is endothermic and hence the actual temperature of the reaction vessel (isolated from the surrounding) may be different from that expected. Given that the initial temperature of the reaction vessel was used in the calculations, how would, this affect the predicted value of moles of $Cl_2$ (n), according to equation: n = PV/RT

Prashant Kotian · Accepted Answer

Correct answer: B. Since the reaction is endothermic and the vessel is isolated (adiabatic), the reaction absorbs heat from the system itself, causing the temperature of the vessel to decrease. Thus, the actual final temperature (T_actual) is less than the initial temperature (T_initial) used in the calculations. In the formula n = PV/RT, if we use a higher T (T_initial instead of the lower T_actual) in the denominator, the calculated or predicted value of moles (n_predicted) will be less than the actual value (n_actual).

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