Applying Hess’s Law What is the enthalpy of reaction, ΔH, for the formation of tungsten carbide, WC, from the elements? (Tungsten carbide is very hard and is used to make cutting tools and rock drills.) W(s) + C(graphite) → WC(s) The enthalpy change for this reaction is difficult to measure

Question

Applying Hess’s Law

What is the enthalpy of reaction, ΔH, for the formation of tungsten carbide, WC, from the elements? (Tungsten carbide is very hard and is used to make cutting tools and rock drills.)

W(s) + C(graphite) → WC(s)

The enthalpy change for this reaction is difficult to measure directly, because the reaction occurs at 1400°C. However, the heats of combustion of the elements and of tungsten carbide can be measured easily:

[latex]\begin{array}{c}2 W (s)+3 O _2(g) \longrightarrow 2 WO _3(s) ; \Delta H=-1685.8 kJ (1)\C ( ext { graphite })+ O _2(g) \longrightarrow CO _2(g) ; \Delta H=-393.5 kJ (2)\ 2 WC (s)+5 O _2(g) \longrightarrow 2 WO _3(s)+2 CO _2(g) ; \Delta H=-2391.8 kJ (3)\end{array}[/latex]

PROBLEM STRATEGY

You need to multiply Equations 1, 2, and 3 by factors so that when you add the three equations you obtain the desired equation for the formation of WC(s). To obtain these factors, compare Equations 1, 2, and 3 in turn with the desired equation. For instance, note that Equation 1 has 2W(s) on the left side, whereas the desired equation has W(s). Therefore, you multiply Equation 1 (and its ΔH) by [latex]\frac{1}{2}[/latex].

Accepted Answer

Multiplying Equation 1 by [latex]\frac{1}{2}[/latex] , you obtain

[latex]W (s)+\frac{3}{2} O _2( g ) \longrightarrow WO _3(s) ; \Delta H=\frac{1}{2} imes(-1685.8 kJ )=-842.9 kJ[/latex]

Compare Equation 2 with the desired equation. Both have C(graphite) on the left side; therefore, you leave Equation 2 as it is. Now, compare Equation 3 with the desired equation. Equation 3 has 2WC(s) on the left side, whereas the desired equation has WC(s) on the right side. Hence, you reverse Equation 3 and multiply it (and its ΔH) by [latex]\frac{1}{2}[/latex] .

[latex]WO _3(s)+ CO _2(g) \longrightarrow WC (s)+\frac{5}{2} O _2(g) ; \Delta H=-\frac{1}{2} imes(-2391.8 kJ )=1195.9 kJ[/latex]

Note that the ΔH is obtained by multiplying the value for Equation 3 by [latex]-\frac{1}{2}[/latex]. Now these three equations and the corresponding ΔH’s are added together.

[latex]W (s)+\cancel{\frac{3}{2} O _2(g)} \quad \longrightarrow \cancel{WO _3(s)} \quad \Delta H=-842.9 kJ\ C ( ext { graphite }) \cancel{O _2(g)} \longrightarrow \cancel{CO _2(g)} \quad \Delta H=-393.5 kJ\ \underline{\cancel{WO _3(s)}+ \cancel{CO _2(g)} \quad \longrightarrow WC (s)+\cancel{\frac{5}{2} O _2(g)}} \quad \underline{\Delta H=1195.9 kJ}\ W (s)+ C ext { (graphite) } \longrightarrow WC (s) \quad \Delta H =- 4 0 . 5 k J[/latex]

Question 6.7: Applying Hess’s Law What is the enthalpy of reaction, ΔH, fo...

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