Question:
What is the work done in the formation of 1 mole gaseous methane from solid carbon and gaseous hydrogen at \(27^{\circ }\text{C}\)? (\(R=8.314\ \text{J k}^{-1}\text{ mol}^{-1}\))
What is the work done in the formation of 1 mole gaseous methane from solid carbon and gaseous hydrogen at \(27^{\circ }\text{C}\)? (\(R=8.314\ \text{J k}^{-1}\text{ mol}^{-1}\))
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Find the change in moles of gas for the formation reaction and use w = -Δn RT.
Updated On: Nov 12, 2026
- \(-2.4942\) kJ
- \(+2.4942\) kJ
- \(-24.942\) kJ
- \(+24.942\) kJ
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
For a reaction at constant temperature and pressure, work done on the system is \(w=-\Delta n_g RT\), where \(\Delta n_g\) is the change in moles of gas.
Step 2: Write the reaction
\(\text{C(s)}+2\text{H}_{2}\text{(g)}\rightarrow\text{CH}_{4}\text{(g)}\)
\(\Delta n_g=1-2=-1\).
Step 3: Calculation
\(T=27+273=300\) K.
\(w=-(-1)\times8.314\times300=+2494.2\) J \(=+2.4942\) kJ.
Final Answer:
The work done is \(+2.4942\) kJ. \[ \boxed{\text{(B) }+2.4942\ \text{kJ}} \]
For a reaction at constant temperature and pressure, work done on the system is \(w=-\Delta n_g RT\), where \(\Delta n_g\) is the change in moles of gas.
Step 2: Write the reaction
\(\text{C(s)}+2\text{H}_{2}\text{(g)}\rightarrow\text{CH}_{4}\text{(g)}\)
\(\Delta n_g=1-2=-1\).
Step 3: Calculation
\(T=27+273=300\) K.
\(w=-(-1)\times8.314\times300=+2494.2\) J \(=+2.4942\) kJ.
Final Answer:
The work done is \(+2.4942\) kJ. \[ \boxed{\text{(B) }+2.4942\ \text{kJ}} \]
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