Question

In the presence of excess oxygen, methane gas burns in a constant-pressure system Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.to yield carbon dioxide and water: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH = -890.0 kJ Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.Calculate the value of q (kJ) in this exothermic reaction when 1.90 g of methane is combusted at constant pressure.

Answer 1

Answer: The value of q for the given reaction is -105.9 kJ

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of methane = 1.90 g

Molar mass of methane = 16 g/mol

Putting values in above equation, we get:

$\text{Moles of methane}=\frac{1.90g}{16g/mol}=0.119mol$

For the given chemical equation:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-890.0kJ$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.119 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -890.0 kJ/mol

Putting values in above equation, we get:

$-890kJ/mol=\frac{q}{0.119mol}\\\\q=(-890kJ/mol\times 0.119mol)=-105.91kJ$

Hence, the value of q for the given reaction is -105.9 kJ