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**Problem**: Given the values of ΔGf° given below in kJ/mol calculate the value of ΔG° in kJ for the reaction,NH3(g) + HBr(g) => NH4Br(s)ΔGf° (NH3(g)) = -17ΔGf° (HBr(g)) = -52ΔGf° (NH4Br(s)) = -177a. 108 kJb. 212 kJc. -108 kJd. -212 kJe. -246 kJ

###### FREE Expert Solution

We’re being asked to calculate the value of ΔG° in kJ for the given reaction:

NH_{3}(g) + HBr(g) → NH_{4}Br(s)

We can use the following equation to solve for ** ΔG˚_{rxn}**:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{prod}}{\mathbf{-}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{react}}}$

The values for **ΔG˚ _{f}** can be looked up in textbooks or online:

###### Problem Details

Given the values of ΔG_{f}° given below in kJ/mol calculate the value of ΔG° in kJ for the reaction,

NH_{3}(g) + HBr(g) => NH_{4}Br(s)

ΔG_{f}° (NH_{3}(g)) = -17

ΔG_{f}° (HBr(g)) = -52

ΔG_{f}° (NH_{4}Br(s)) = -177

a. 108 kJ

b. 212 kJ

c. -108 kJ

d. -212 kJ

e. -246 kJ

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