Question

Consider the following reaction:

Answer 1

Answer:

(a) $\Delta H=-180.7kJ$

(b) $\Delta H=-6.48x10^4kJ$

Explanation:

Hello.

(a) in this case, since the change in the molar enthalpy for this reaction is ΔH= -2891 kJ per mole of methane, we can compute the total change in the enthalpy by considering the given mass of methane which should be converted to moles as shown below:

$\Delta H=1.00gCH_4*\frac{1molCH_4}{16gCH_4} *-2891\frac{kJ}{molCH_4} \\\\\Delta H=-180.7kJ$

(b) For this one, we first compute the moles of methane by using the ideal gas equation:

$n=\frac{PV}{RT}=\frac{740torr*\frac{1atm}{760torr}*1.003x10^3L}{0.082\frac{atm*L}{mol*K}*(258+273.15)K}\\ \\n=22.4mol$

Then, we compute the change in the enthalpy as done on (a):

$\Delta H=22.4mol*-2891\frac{kJ}{molCH_4} \\\\\Delta H=-6.48x10^4kJ$

Regards.