Question

Strontium metal is responsible for the red color in fireworks. Fireworks manufacturers use strontium carbonate, which can be produced by combining strontium metal, graphite (C), and oxygen gas. The formation of one mole of \(\mathrm{SrCO}_{3}\) releases \(1.220 \times 10^{3} \mathrm{~kJ}\) of energy. (a) Write a balanced thermochemical equation for the reaction. (b) What is \(\Delta H\) when \(10.00 \mathrm{~L}\) of oxygen at \(25^{\circ} \mathrm{C}\) and 1.00 atm are used by the reaction?

Short answer

Answer: The enthalpy change for the reaction with the given oxygen is approximately -331.85 kJ.

Step-by-step solution

Write the unbalanced equation

The reaction occurs between strontium metal (Sr), graphite (C), and oxygen gas (O2) to form strontium carbonate (SrCO3). Therefore, the unbalanced equation is: $$ \mathrm{Sr}+\mathrm{C}+\mathrm{O}_{2}\rightarrow\mathrm{SrCO}_{3} $$

Balance the equation and include the enthalpy change

To balance the equation, we need one Sr, one C, and 3 O atoms on both sides: $$ \mathrm{Sr}+\mathrm{C}+\frac{3}{2}\mathrm{O}_{2}\rightarrow\mathrm{SrCO}_{3} $$ Now, we include the energy released per mole of SrCO3 in the equation: $$ \mathrm{Sr}+\mathrm{C}+\frac{3}{2}\mathrm{O}_{2}\rightarrow\mathrm{SrCO}_{3} \quad \Delta H=-1.220 \times 10^{3} \,\mathrm{kJ} $$

Calculate the moles of oxygen at 25°C and 1 atm

We are given the volume of oxygen as 10.00 L at 25°C and 1 atm. We can use the ideal gas law to find the moles of oxygen: $$ PV = nRT $$ Where P = pressure in atm, V = volume in L, n = moles, R = gas constant (0.0821 L atm/mol K), and T = temperature in K. We'll first convert temperature to Kelvin: $$ T(\mathrm{K})=T(^{\circ}\mathrm{C})+273.15=25+273.15=298.15\,\mathrm{K} $$ Now, we can find the moles of oxygen: $$ n=\frac{PV}{RT}=\frac{(1\,\mathrm{atm})(10\,\mathrm{L})}{(0.0821\,\mathrm{L\,atm/mol\,K})(298.15\,\mathrm{K}} \approx0.408\,\mathrm{mol} $$

Calculate the moles of SrCO3 that can be produced

From the balanced equation, for every 3/2 moles of O2, we get 1 mole of SrCO3. Therefore, the moles of SrCO3 produced using 0.408 mol of O2 can be calculated as follows: $$ \mathrm{moles\,of\,SrCO}_{3}=\mathrm{moles\,of\,O}_{2}\times\frac{1\,\mathrm{mol\,SrCO}_{3}}{\frac{3}{2}\,\mathrm{mol\,O}_{2}}\approx0.272\,\mathrm{mol} $$

Find the enthalpy change for the reaction with the given oxygen

Now, we can find the enthalpy change using the given energy per mole of SrCO3 and the moles of SrCO3 calculated in the previous step: $$ \Delta H (\mathrm{kJ})= \mathrm{moles\,of\,SrCO}_{3}\times\Delta H_{\mathrm{per\,mol}} = (0.272\,\mathrm{mol})(-1.220\times10^{3}\,\mathrm{kJ/mol}) \approx -331.85\,\mathrm{kJ} $$ So, the enthalpy change for the reaction with the given oxygen is approximately -331.85 kJ.