In week five, Aaron Jones recorded four touchdowns in the Green Bay Packers’ big win over the Dallas Cowboys. In week seven, it was a six-touchdown game from quarterback Aaron Rodgers that propelled the Packers to victory over the Oakland Raiders.
Those two performances now have something in common: they both earned an Aaron the title of NFC Offensive Player of the Week. Rodgers was awarded that honor on Wednesday for his game on Sunday, when he put up career-best numbers without his best wide receiver.
There was really no other choice for the honor this week. Rodgers posted the first perfect passer rating in a full game for his career, set a career-high in single-game yards per attempt with 13.84, and put up his ninth game with at least 80 percent of his passes going for completions. His final stat line ran 25-for-31 for 429 yards and five passing scores, with two rushing attempts for six yards and another touchdown.
As an added bonus, Rodgers threw his 18th career touchdown of 70 or more yards in the game, a 74-yard catch-and-run by Marquez Valdes-Scantling. Incredibly, this was only the second such play since the 2014 season, joining last year’s 75-yard game-winner against the Chicago Bears.
The last time Rodgers earned the NFC’s Offensive Player of the Week honor was against the Dallas Cowboys in 2017, when he threw three touchdowns including a game-winner to Davante Adams with just a few seconds remaining in the game. This is also the Packers’ third Player of the Week honor in 2019; Jones earned the honor on offense in week five while outside linebacker Preston Smith was named defensive player of the week for his week three game against the Denver Broncos, when he recorded three sacks and forced a fumble.