In the equilibrium reaction `N_(2)+3H_(2) Leftrightarrow 2NH_(3)`, the sign of `triangleH` accompanying the reaction is

Options

A. Positive
B. Negative
C. May be positive or negative
D. Cannot be predicted

Answer

To determine the sign of ΔH (delta H) for the equilibrium reaction \( N_2 + 3H_2 \leftrightarrow 2NH_3 \), we can follow these steps:

### Step 1: Identify the Reaction Type
The reaction given is the synthesis of ammonia from nitrogen and hydrogen. This is a well-known reaction called the Haber process.

**Hint:** Recognize that the Haber process is a key industrial reaction for producing ammonia.

### Step 2: Understand the Nature of the Reaction
In this reaction, one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia. This process involves the formation of chemical bonds in ammonia, which releases energy.

**Hint:** Consider the bond formation and breaking involved in the reaction.

### Step 3: Determine the Energy Change
The formation of ammonia from nitrogen and hydrogen is an exothermic reaction. This means that energy is released when ammonia is formed. The standard enthalpy change (ΔH) for this reaction is approximately -91.8 kJ/mol, indicating that heat is released.

**Hint:** Recall that exothermic reactions have a negative ΔH value.

### Step 4: Conclude the Sign of ΔH
Since the reaction releases energy, the sign of ΔH for the reaction \( N_2 + 3H_2 \leftrightarrow 2NH_3 \) is negative.

**Hint:** Remember that a negative ΔH signifies an exothermic reaction.

### Final Answer
The sign of ΔH accompanying the reaction \( N_2 + 3H_2 \leftrightarrow 2NH_3 \) is **negative**.

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