NFL Rookie Of The Year Award Winner 2019: Saquon Barkley Earns Honor

Former Penn State Nittany Lions star running back Saquon Barkley entered the NFL with high expectations. During his collegiate career, Barkley rushed for 3,843 yards with 43 rushing touchdowns, and added 1,195 receiving yards with 8 touchdown receptions.
Barkley was ultimately drafted with the No. 2 overall pick in the 2018 NFL Draft by the New York Giants.
Now, he has picked up the highest honor for a first-year player in the league.
According to reports, Barkley has been named the 2018 Pepsi NFL Rookie of the Year.
“Saquon Barkley won the Pepsi Rookie of the Year vote, source confirms. Expecting official announcement to come soon, possible today. The Offensive Rookie of the Year from the league will be awarded Saturday night at NFL Honors,” New York Giants reporter Art Stapleton tweeted.

“If you ask my peers that played against me or coaches that played against me and schemed against me who was the Rookie of the Year, I think they would say me,” Barkley said in a recent interview with ESPN.
During his rookie campaign, the former Nittany Lions star rushed for 1,307 yards and 11 touchdowns on the ground. To show off his versatility, Barkley hauled in 91 catches for 721 yards and an additional four touchdowns.
An official announcement will be made on Saturday, February 2.

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