# What is the enthalpy of reaction in #"kJ/mol"# for when #"0.254 g"# of #"Mg"# is dropped into a coffee-cup calorimeter, if the temperature of #"200. mL"# of water rose from #25.00^@ "C"# to #32.20^@ "C"#?

##### 1 Answer

I got

Well, the first thing you should recognize is that

Furthermore, by **conservation of energy**, we assume that **released by** the reaction is completely **absorbed into** the calorimeter solution to increase its temperature.

(It's not perfect, as the calorimeter is open to the atmosphere to manage a constant pressure, but it's a pretty decent assumption.)

From that, we have the main equation:

#bb(q_"cal" = m_wc_wDeltaT)# where:

#m_w# is the mass of the solution, obtained by using the volume of the product solution and the density of the solution (which we assume is that of pure water).#c_w = "4.184 J/g"^@ "C"# is the specific heat capacity of the solution, again assumed to be equal to that for pure water.#DeltaT# is the change in temperature of the solution due to the exothermic reaction.

So, the main steps would be:

- Find
#q_"cal"# , in#"kJ"# . - Solve for
#q_"rxn"# via the sign difference, in#"kJ"# . - Solve for
#DeltaH_"rxn"# in#"kJ""/"ul(ul("mol"))# by utilizing the given mass of#"Mg"(s)# .

**FINDING HEAT EVOLVED IN THE CALORIMETER**

#color(green)(q_"cal") = ("200. mL" xx "1.00 g"/"mL")("4.184 J/g"^@ "C")(32.20^@ "C" - 25.00^@ "C")#

#=# #6.02_(496)xx10^3# #"J"#

#= color(green)(6.02_(496))# #color(green)("kJ")#

where subscripts indicate the uncertain digits (the digits past the last significant figure).

**FINDING HEAT EVOLVED DUE TO THE REACTION**

This is easy. We stated that

#color(green)(q_"rxn" = -6.02_(496))# #color(green)("kJ")#

**FINDING ENTHALPY OF THE REACTION**

This part is where you really have to think. The reaction is based on the given mass of

The

#color(green)(n_("Mg"(s))) = "0.254 g Mg"(s) xx ("1 mol Mg"(s))/("24.305 g Mg"(s))#

#= color(green)(0.0104_(505))# #color(green)("mols Mg"(s))#

Finally, we note the unit conversion to get:

#color(blue)(DeltaH_"rxn") = q_"rxn"/(n_("Mg"(s)))#

#= -(6.02_(496) "kJ")/(0.0104_(505) "mols Mg"(s))#

#= color(blue)(-"577 kJ/mol")# (rounded to three sig figs)