I have to calculate the ΔV= final volume-initial volume
For the first trial the final volume was .005mL the initial volume was .03mL, which equaled to
-.025mL. The volume was negative because it was exothermic reaction. For the second trial the final volume was 0.015 and the initial volume was 0.03, which equaled to -.015mL.
Now I have to convert ΔV to ΔH273 values. TheΔH273 is the molar enthalpy change for the reaction of the magnesium metal. How do I compute the mean value of ΔH273?
ΔH=ΔE +nRT
Also calculate the ΔE273 from the mean value of ΔH273?
The equation for the reaction that took place was: Mg(s) +2H+(aq)→Mg+2(aq) +H2 (aq)
To compute the mean value of ΔH273, you will need to use the equation ΔH = ΔE + nRT, where ΔH is the enthalpy change, ΔE is the internal energy change, n is the number of moles of the reacting substance, R is the ideal gas constant, and T is the temperature.
First, calculate the ΔH273 for each trial using the formula:
ΔH273 = ΔE + nRT
Since the reaction is exothermic, the value of ΔE will be negative. You have already calculated the values of ΔV for each trial, which are -0.025 mL and -0.015 mL, representing the change in volume. However, you need to convert these values to liters by dividing by 1000 to get -0.000025 L and -0.000015 L.
Next, calculate the number of moles of magnesium (Mg) used in each trial. The balanced equation tells us that 1 mole of Mg reacts with 2 moles of H+, so the number of moles of Mg is half the number of moles of H+.
Assuming the concentration of H+ remains constant between the trials, you can calculate the number of moles of Mg from the change in volume of H+. The change in volume of H+ is equal to the change in volume of the reaction, ΔV.
In the first trial, ΔV is -0.000025 L, so the number of moles of Mg is 0.5 times the absolute value of ΔV divided by the volume of H+ solution. The initial volume of H+ is 0.03 mL, which is 0.00003 L. Therefore, the number of moles of Mg is 0.5 * 0.000025 L / 0.00003 L = 0.000020833 moles.
Repeat the same calculation for the second trial using the values of -0.000015 L for ΔV and 0.03 mL for the initial volume of H+ solution.
Once you have the values of ΔH273 for each trial, take the mean value by averaging the two values.
To calculate the ΔE273 from the mean value of ΔH273, you can rearrange the equation ΔH = ΔE + nRT. Given that ΔH273 is the molar enthalpy change for the reaction and you know the values of n, R, and T (273 K), you can solve for ΔE273 by rearranging the equation as follows:
ΔE273 = ΔH273 - nRT
Substitute the mean value of ΔH273 and the appropriate values for n, R, and T into the equation and calculate the value of ΔE273.