Question

in problems 4 - 7, calculate ah for the reaction:
3 ch₄(g) → c₃h₈(g) + 2 h₂(g)

1. using the thermochemical equations

2 c₃h₈(g) + 10 o₂(g) → 6 co₂(g) + 8 h₂o(l)
+ ch₄(g) + 2 o₂(g) → co₂(g) + 2 h₂o(l)
+ 2 h₂(g) + o₂(g) → 2 h₂o(l)
δh = - 4441 kj
δh = - 890 kj
δh = - 572 kj

1. given that when, in the reaction above, 2.000 g of ch₄(g) is used in a calorimeter

with heat capacity of 2139 j/°c. the temperature falls from 25.26 to 22.88°c.

Explanation:

Step1: Calculate the temperature change

The temperature change \(\Delta T=T_{final}-T_{initial}\). Given \(T_{initial} = 25.26^{\circ}C\) and \(T_{final}=22.88^{\circ}C\), so \(\Delta T=22.88 - 25.26=- 2.38^{\circ}C\)

Step2: Use the heat - capacity formula

The heat \(q = C\times\Delta T\), where \(C = 2139J/^{\circ}C\) and \(\Delta T=-2.38^{\circ}C\). Then \(q=2139\times(- 2.38)=-5090.82J\approx - 5.09kJ\)

Step3: Calculate the number of moles of \(CH_{4}\)

The molar mass of \(CH_{4}\) is \(M=(12 + 4\times1)g/mol = 16g/mol\). Given \(m = 2.000g\), the number of moles \(n=\frac{m}{M}=\frac{2.000g}{16g/mol}=0.125mol\)

Step4: Calculate the enthalpy change per mole

If \(0.125mol\) of \(CH_{4}\) gives off \(q=-5.09kJ\) (exothermic, so \(\Delta H\) is negative), then for \(1mol\) of \(CH_{4}\), \(\Delta H=\frac{-5.09kJ}{0.125mol}=-40.72kJ/mol\)

Answer:

\(-40.72kJ/mol\)